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And we are going to be left with negative four y plus three y. Well, that's just going to be negative y. So by adding the left-hand side of this bottom equation to the left-hand side of the top equation, we were able to cancel out the x's. We had x and we had a negative x. That was very nice for us. So what do we do on the right-hand side? We've already said that we have to add the same thing to both sides of an equation. | Solving system with elimination Algebra Khan Academy.mp3 |
We had x and we had a negative x. That was very nice for us. So what do we do on the right-hand side? We've already said that we have to add the same thing to both sides of an equation. We might be tempted, we might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative x plus three y to that side. But that's not going to help us much. We're going to have negative 18 minus x plus three y. | Solving system with elimination Algebra Khan Academy.mp3 |
We've already said that we have to add the same thing to both sides of an equation. We might be tempted, we might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative x plus three y to that side. But that's not going to help us much. We're going to have negative 18 minus x plus three y. We would have introduced an x on the right-hand side of the equation. But what if we could add something that's equivalent to negative x plus three y that does not introduce the x variable? Well, we know that the number 11 is equivalent to negative x plus three y. | Solving system with elimination Algebra Khan Academy.mp3 |
We're going to have negative 18 minus x plus three y. We would have introduced an x on the right-hand side of the equation. But what if we could add something that's equivalent to negative x plus three y that does not introduce the x variable? Well, we know that the number 11 is equivalent to negative x plus three y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. | Solving system with elimination Algebra Khan Academy.mp3 |
Well, we know that the number 11 is equivalent to negative x plus three y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative x plus three y. But the second equation tells us that negative x plus three y is going to be equal to 11. It's introducing that second constraint. | Solving system with elimination Algebra Khan Academy.mp3 |
So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative x plus three y. But the second equation tells us that negative x plus three y is going to be equal to 11. It's introducing that second constraint. And so let's add 11 to the right-hand side, which is, once again, I know I keep repeating it, it's the same thing as negative x plus three y. So negative 18 plus 11 is negative seven. And since we added the same thing to both sides, the equality still holds. | Solving system with elimination Algebra Khan Academy.mp3 |
It's introducing that second constraint. And so let's add 11 to the right-hand side, which is, once again, I know I keep repeating it, it's the same thing as negative x plus three y. So negative 18 plus 11 is negative seven. And since we added the same thing to both sides, the equality still holds. And we get negative y is equal to negative seven. Or divide both sides by negative one or multiply both sides by negative one. So multiply both sides by negative one. | Solving system with elimination Algebra Khan Academy.mp3 |
And since we added the same thing to both sides, the equality still holds. And we get negative y is equal to negative seven. Or divide both sides by negative one or multiply both sides by negative one. So multiply both sides by negative one. We get y is equal to seven. So we have the y coordinate of the xy pair that satisfies both of these. Now how do we find the x? | Solving system with elimination Algebra Khan Academy.mp3 |
So multiply both sides by negative one. We get y is equal to seven. So we have the y coordinate of the xy pair that satisfies both of these. Now how do we find the x? Well, we can just substitute this y equals seven to either one of these. When y equals seven, we should get the same x regardless of which equation we use. So let's use the top equation. | Solving system with elimination Algebra Khan Academy.mp3 |
Now how do we find the x? Well, we can just substitute this y equals seven to either one of these. When y equals seven, we should get the same x regardless of which equation we use. So let's use the top equation. So we know that x minus four times, instead of writing y, I'm gonna write four times seven, because we're gonna figure out what is x when y is seven. That is going to be equal to negative 18. And so let's see, negative four, or four times seven, that is 28. | Solving system with elimination Algebra Khan Academy.mp3 |
So let's use the top equation. So we know that x minus four times, instead of writing y, I'm gonna write four times seven, because we're gonna figure out what is x when y is seven. That is going to be equal to negative 18. And so let's see, negative four, or four times seven, that is 28. So let's see, I could, to solve for x, I could add 28 to both sides. So add 28 to both sides. On the left-hand side, negative 28, positive 28, those cancel out, I'm just left with an x. | Solving system with elimination Algebra Khan Academy.mp3 |
And so let's see, negative four, or four times seven, that is 28. So let's see, I could, to solve for x, I could add 28 to both sides. So add 28 to both sides. On the left-hand side, negative 28, positive 28, those cancel out, I'm just left with an x. And on the right-hand side, I get negative eight, negative 18 plus 28 is 10. So there you have it. I have the xy pair that satisfies both. | Solving system with elimination Algebra Khan Academy.mp3 |
On the left-hand side, negative 28, positive 28, those cancel out, I'm just left with an x. And on the right-hand side, I get negative eight, negative 18 plus 28 is 10. So there you have it. I have the xy pair that satisfies both. X equals 10, y equals seven. I could write it here. So I could write it as coordinates. | Solving system with elimination Algebra Khan Academy.mp3 |
I have the xy pair that satisfies both. X equals 10, y equals seven. I could write it here. So I could write it as coordinates. I could write it as 10 comma seven. And notice, what I just did here, I encourage you to substitute y equals seven here, and you will also get x equals 10. Either way, you would have come to x equals 10. | Solving system with elimination Algebra Khan Academy.mp3 |
So I could write it as coordinates. I could write it as 10 comma seven. And notice, what I just did here, I encourage you to substitute y equals seven here, and you will also get x equals 10. Either way, you would have come to x equals 10. And to visualize what is going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant to draw a straighter line than that. | Solving system with elimination Algebra Khan Academy.mp3 |
Either way, you would have come to x equals 10. And to visualize what is going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant to draw a straighter line than that. All right, there you go. So let's say that is our y-axis, and that is, whoops, that is our x-axis. And then, let's see, the top equation is gonna look something like this. | Solving system with elimination Algebra Khan Academy.mp3 |
Whoops, I meant to draw a straighter line than that. All right, there you go. So let's say that is our y-axis, and that is, whoops, that is our x-axis. And then, let's see, the top equation is gonna look something like this. It's gonna look something like this. And then that bottom equation is gonna look something, something like, let me draw it a little bit nicer than that. It's gonna look something like this, something like that. | Solving system with elimination Algebra Khan Academy.mp3 |
And then, let's see, the top equation is gonna look something like this. It's gonna look something like this. And then that bottom equation is gonna look something, something like, let me draw it a little bit nicer than that. It's gonna look something like this, something like that. Let me draw that bottom one here so you see the point of intersection. And so the point of intersection right over here, that is an xy pair that satisfies both of these equations. And that, we just saw, it happens when x is equal to 10 and y is equal to seven. | Solving system with elimination Algebra Khan Academy.mp3 |
It's gonna look something like this, something like that. Let me draw that bottom one here so you see the point of intersection. And so the point of intersection right over here, that is an xy pair that satisfies both of these equations. And that, we just saw, it happens when x is equal to 10 and y is equal to seven. Once again, this white line, that's all the x and y pairs that satisfy the top equation. This orange line, that's all the x and y pairs that satisfy the orange equation. And where they intersect, that point is on both lines. | Solving system with elimination Algebra Khan Academy.mp3 |
We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now, if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side. And then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So let's just do that. So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So we can factor that out. So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
s plus 5 times s would give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or. It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
Actually, we could say and or. It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared. x times b is bx. a times x is plus ax. a times b is ab. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
x times x is x squared. x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here. We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
We have 1 as a leading coefficient here. We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point to, well actually this was a case of s. So we could have factored straight to the case of s plus 5 times s minus 7. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point to, well actually this was a case of s. So we could have factored straight to the case of s plus 5 times s minus 7. We could have done that straight away and we've gotten to that right there. And of course, that whole thing was equal to 0. So that would have been a little bit of a shortcut, but factoring by grouping, it's a completely appropriate way to do it as well. | Solving a quadratic equation by factoring Algebra II Khan Academy.mp3 |
And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any line that describes how steep it is, how quickly does it increase or decrease. So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well, one way to think about it could say, well, how much does a line increase for, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we, if an increase in vertical for a given increase in horizontal, for a given increase in horizontal. | Introduction to slope Algebra I Khan Academy.mp3 |
What's a reasonable way to assign a number to these lines that describe their steepness? Well, one way to think about it could say, well, how much does a line increase for, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we, if an increase in vertical for a given increase in horizontal, for a given increase in horizontal. So how could this give us a value? Well, let's look at that magenta line again. Now, let's just start at an arbitrary point to that magenta line, but I'll start at a point where it's gonna be easy for me to figure out what point we're at. | Introduction to slope Algebra I Khan Academy.mp3 |
So let's say if we, if an increase in vertical for a given increase in horizontal, for a given increase in horizontal. So how could this give us a value? Well, let's look at that magenta line again. Now, let's just start at an arbitrary point to that magenta line, but I'll start at a point where it's gonna be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one, so I move one to the right, to get back on the line, how much do I have to increase in the vertical direction? Well, I have to increase in the vertical direction by two, by two. So at least for this magenta line, it looks like our increase in vertical is two. | Introduction to slope Algebra I Khan Academy.mp3 |
Now, let's just start at an arbitrary point to that magenta line, but I'll start at a point where it's gonna be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one, so I move one to the right, to get back on the line, how much do I have to increase in the vertical direction? Well, I have to increase in the vertical direction by two, by two. So at least for this magenta line, it looks like our increase in vertical is two. Whenever we have an increase in one in the horizontal direction. And we could, let's see, does that apply? Let's see, does that still work? | Introduction to slope Algebra I Khan Academy.mp3 |
So at least for this magenta line, it looks like our increase in vertical is two. Whenever we have an increase in one in the horizontal direction. And we could, let's see, does that apply? Let's see, does that still work? If I were to go, if I were to start here, instead of increasing the horizontal direction by one, if I were to increase in the horizontal direction, let's increase by three. So now, I've gone plus three in the horizontal direction. Then to get back on the line, how much do I have to increase in the vertical direction? | Introduction to slope Algebra I Khan Academy.mp3 |
Let's see, does that still work? If I were to go, if I were to start here, instead of increasing the horizontal direction by one, if I were to increase in the horizontal direction, let's increase by three. So now, I've gone plus three in the horizontal direction. Then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six. I have to increase by six, so plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. | Introduction to slope Algebra I Khan Academy.mp3 |
Then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six. I have to increase by six, so plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We're just saying, hey, let's just measure how much we increase in vertical for a given increase in horizontal. Well, two over one is just two, and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm gonna increase twice as much, twice as much in the vertical direction, twice as much in the vertical direction. | Introduction to slope Algebra I Khan Academy.mp3 |
So when I increase by three in the horizontal direction, I increase by six in the vertical. We're just saying, hey, let's just measure how much we increase in vertical for a given increase in horizontal. Well, two over one is just two, and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm gonna increase twice as much, twice as much in the vertical direction, twice as much in the vertical direction. So this notion of increase in vertical divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines, and this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for, say, a ski slope, and that's because a ski slope has a certain inclination. | Introduction to slope Algebra I Khan Academy.mp3 |
So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm gonna increase twice as much, twice as much in the vertical direction, twice as much in the vertical direction. So this notion of increase in vertical divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines, and this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for, say, a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is, and the convention is we measure the increase in vertical for a given increase in horizontal. So two over one is equal to six over three is equal to two. | Introduction to slope Algebra I Khan Academy.mp3 |
And you're probably familiar with the notion of the word slope being used for, say, a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is, and the convention is we measure the increase in vertical for a given increase in horizontal. So two over one is equal to six over three is equal to two. This is equal to the slope of this magenta line. So slope, so let me write this down. So this slope right over here, the slope of that line is going to be equal to two. | Introduction to slope Algebra I Khan Academy.mp3 |
So two over one is equal to six over three is equal to two. This is equal to the slope of this magenta line. So slope, so let me write this down. So this slope right over here, the slope of that line is going to be equal to two. And one way to interpret that, if you, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. Now what about this blue line here? What would be the slope of the blue line? | Introduction to slope Algebra I Khan Academy.mp3 |
So this slope right over here, the slope of that line is going to be equal to two. And one way to interpret that, if you, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. Now what about this blue line here? What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope, and this is just the convention that mathematicians have defined for slope, but it's a valuable one. What is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. | Introduction to slope Algebra I Khan Academy.mp3 |
What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope, and this is just the convention that mathematicians have defined for slope, but it's a valuable one. What is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So change in vertical, and in this coordinate it's going to be, the vertical is our y-coordinate, divided by our change in horizontal. And x is our horizontal coordinate in this coordinate plane right over here. And you might say, wait, wait, wait, you said change in, but then you drew this triangle. | Introduction to slope Algebra I Khan Academy.mp3 |
And I'll introduce a new notation for you. So change in vertical, and in this coordinate it's going to be, the vertical is our y-coordinate, divided by our change in horizontal. And x is our horizontal coordinate in this coordinate plane right over here. And you might say, wait, wait, wait, you said change in, but then you drew this triangle. Well, this triangle, this is the Greek letter delta. This is the Greek letter delta, and it's a math symbol used to represent change in. So that's delta, delta. | Introduction to slope Algebra I Khan Academy.mp3 |
And you might say, wait, wait, wait, you said change in, but then you drew this triangle. Well, this triangle, this is the Greek letter delta. This is the Greek letter delta, and it's a math symbol used to represent change in. So that's delta, delta. And it literally means change in y, change in y, divided by change in x. Change in x. So if we want to find the slope of the blue line, we just have to say, well, how much does y change for a given change in x? | Introduction to slope Algebra I Khan Academy.mp3 |
So that's delta, delta. And it literally means change in y, change in y, divided by change in x. Change in x. So if we want to find the slope of the blue line, we just have to say, well, how much does y change for a given change in x? So the slope of the blue line. So let's see, let's see, actually let me do this, let me do it this way. Let me have a, when if I, let's just start at some point here, and let's say if my x changes by two, so my delta x is equal to positive two, what's my delta y going to be? | Introduction to slope Algebra I Khan Academy.mp3 |
So if we want to find the slope of the blue line, we just have to say, well, how much does y change for a given change in x? So the slope of the blue line. So let's see, let's see, actually let me do this, let me do it this way. Let me have a, when if I, let's just start at some point here, and let's say if my x changes by two, so my delta x is equal to positive two, what's my delta y going to be? What's going to be my change in y? Well, if I go to the right by two, to get back on the line, I have to increase my y by two. So my change in y is also going to be plus two. | Introduction to slope Algebra I Khan Academy.mp3 |
Let me have a, when if I, let's just start at some point here, and let's say if my x changes by two, so my delta x is equal to positive two, what's my delta y going to be? What's going to be my change in y? Well, if I go to the right by two, to get back on the line, I have to increase my y by two. So my change in y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in y over change in x, we just saw that when our change in x is positive two, our change in y is also positive two. So our slope is two divided by two, which is equal to one. Which tells us, however much we increase in x, we're going to increase that same amount in y. | Introduction to slope Algebra I Khan Academy.mp3 |
So my change in y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in y over change in x, we just saw that when our change in x is positive two, our change in y is also positive two. So our slope is two divided by two, which is equal to one. Which tells us, however much we increase in x, we're going to increase that same amount in y. And you see that, you increase one in x, you increase one in y. Increase one in x, increase one in y. From any point on the line, that's going to be true. | Introduction to slope Algebra I Khan Academy.mp3 |
Which tells us, however much we increase in x, we're going to increase that same amount in y. And you see that, you increase one in x, you increase one in y. Increase one in x, increase one in y. From any point on the line, that's going to be true. You increase three in x, you're going to increase three in y. It's actually true the other way. If you decrease one in x, you're going to decrease one in y. | Introduction to slope Algebra I Khan Academy.mp3 |
From any point on the line, that's going to be true. You increase three in x, you're going to increase three in y. It's actually true the other way. If you decrease one in x, you're going to decrease one in y. If you decrease two in x, you're going to decrease two in y. And that makes sense from the math of it as well. Because if your change in x is negative two, that's what we did right over here, our change in x was negative two, we went two back, then your change in y is going to be negative two as well. | Introduction to slope Algebra I Khan Academy.mp3 |
And what they've already graphed for us, this right over here, this is the graph of y is equal to the absolute value of x. So let's do this through a series of transformations. So the next thing I wanna graph, let's see if we can graph y, y is equal to the absolute value of x plus three. Now in previous videos, we have talked about it. If you replace your x with an x plus three, this is going to shift your graph to the left by three. You could view this as the same thing as y is equal to the absolute value of x minus negative three. And whatever you are subtracting from this x, that is how much you are shifting it. | Graphing a shifted and stretched absolute value function.mp3 |
Now in previous videos, we have talked about it. If you replace your x with an x plus three, this is going to shift your graph to the left by three. You could view this as the same thing as y is equal to the absolute value of x minus negative three. And whatever you are subtracting from this x, that is how much you are shifting it. So you're going to shift it three to the left. And we're gonna do that right now, and then we're just gonna confirm that it matches up, that it makes sense. So let's first graph that. | Graphing a shifted and stretched absolute value function.mp3 |
And whatever you are subtracting from this x, that is how much you are shifting it. So you're going to shift it three to the left. And we're gonna do that right now, and then we're just gonna confirm that it matches up, that it makes sense. So let's first graph that. I'll get my ruler tool here. So if we shift three to the left, it's gonna look something like, it's gonna look something like this. So on that, when whatever we have inside the absolute value sign is positive, we're going to get essentially this slope of one. | Graphing a shifted and stretched absolute value function.mp3 |
So let's first graph that. I'll get my ruler tool here. So if we shift three to the left, it's gonna look something like, it's gonna look something like this. So on that, when whatever we have inside the absolute value sign is positive, we're going to get essentially this slope of one. And whenever we have inside the absolute value sign is negative, we're gonna have a slope of essentially negative one. So it's gonna look, it's going to look like that. And let's confirm whether this actually makes sense. | Graphing a shifted and stretched absolute value function.mp3 |
So on that, when whatever we have inside the absolute value sign is positive, we're going to get essentially this slope of one. And whenever we have inside the absolute value sign is negative, we're gonna have a slope of essentially negative one. So it's gonna look, it's going to look like that. And let's confirm whether this actually makes sense. Below x equals negative three, for x values less than negative three, what we're gonna have here is this inside of the absolute value sign is going to be negative, and so then we're gonna take its opposite value, and so that makes sense. That's why you have this downward line right over here. Now for x is greater than negative three, when you add three to it, you're going to get a positive value, and so that's why you have this upward sloping line right over here. | Graphing a shifted and stretched absolute value function.mp3 |
And let's confirm whether this actually makes sense. Below x equals negative three, for x values less than negative three, what we're gonna have here is this inside of the absolute value sign is going to be negative, and so then we're gonna take its opposite value, and so that makes sense. That's why you have this downward line right over here. Now for x is greater than negative three, when you add three to it, you're going to get a positive value, and so that's why you have this upward sloping line right over here. And at x equals negative three, you have zero inside the absolute value sign, just as if you didn't shift it, you would have had zero inside the absolute value sign at x equals zero. So hopefully that makes a little bit more intuitive sense of why if you replace x, if you replace x with x plus three, and this isn't just true of absolute value functions, this is true of any function, if you replace x with x plus three, you're going to shift three to the left. All right, now let's keep building. | Graphing a shifted and stretched absolute value function.mp3 |
Now for x is greater than negative three, when you add three to it, you're going to get a positive value, and so that's why you have this upward sloping line right over here. And at x equals negative three, you have zero inside the absolute value sign, just as if you didn't shift it, you would have had zero inside the absolute value sign at x equals zero. So hopefully that makes a little bit more intuitive sense of why if you replace x, if you replace x with x plus three, and this isn't just true of absolute value functions, this is true of any function, if you replace x with x plus three, you're going to shift three to the left. All right, now let's keep building. Now let's see if we can graph y is equal to two times the absolute value of x plus three. So what this is essentially going to do is multiply the slopes by a factor of two. It's going to stretch it vertically by a factor of two. | Graphing a shifted and stretched absolute value function.mp3 |
All right, now let's keep building. Now let's see if we can graph y is equal to two times the absolute value of x plus three. So what this is essentially going to do is multiply the slopes by a factor of two. It's going to stretch it vertically by a factor of two. So for x values greater than negative three, instead of having a slope of one, you're gonna have a slope of two. So let me get my ruler out again and see if I can draw that. So let me put that there. | Graphing a shifted and stretched absolute value function.mp3 |
It's going to stretch it vertically by a factor of two. So for x values greater than negative three, instead of having a slope of one, you're gonna have a slope of two. So let me get my ruler out again and see if I can draw that. So let me put that there. And then, so here instead of a slope of one, I'm gonna have a slope of two. Let me draw that. It's gonna look like that right over there. | Graphing a shifted and stretched absolute value function.mp3 |
So let me put that there. And then, so here instead of a slope of one, I'm gonna have a slope of two. Let me draw that. It's gonna look like that right over there. And then instead of having a slope of negative one for values less than x equals negative three, I'm gonna have a slope of negative two. Let me draw that right over there. So that is the graph of y is equal to two times the absolute value of x plus three. | Graphing a shifted and stretched absolute value function.mp3 |
It's gonna look like that right over there. And then instead of having a slope of negative one for values less than x equals negative three, I'm gonna have a slope of negative two. Let me draw that right over there. So that is the graph of y is equal to two times the absolute value of x plus three. And now to get to the f of x that we care about, we now need to add this two. So now I wanna graph, and we're in the home stretch, I wanna graph y is equal to two times the absolute value of x plus three plus two. Well, whatever y value I was getting for this orange function, I now wanna add two to it. | Graphing a shifted and stretched absolute value function.mp3 |
So that is the graph of y is equal to two times the absolute value of x plus three. And now to get to the f of x that we care about, we now need to add this two. So now I wanna graph, and we're in the home stretch, I wanna graph y is equal to two times the absolute value of x plus three plus two. Well, whatever y value I was getting for this orange function, I now wanna add two to it. So this is just gonna shift it up vertically by two. So instead of, so this is gonna be shifted up by two. This is going to be shifted, every point is going to be shifted up by two. | Graphing a shifted and stretched absolute value function.mp3 |
Well, whatever y value I was getting for this orange function, I now wanna add two to it. So this is just gonna shift it up vertically by two. So instead of, so this is gonna be shifted up by two. This is going to be shifted, every point is going to be shifted up by two. Or you can think about shifting up the entire graph by two. Here in the orange function, whatever y value I got for the black function, I'm gonna have to get two more than that. And so it's going to look, it's going to look like this. | Graphing a shifted and stretched absolute value function.mp3 |
This is going to be shifted, every point is going to be shifted up by two. Or you can think about shifting up the entire graph by two. Here in the orange function, whatever y value I got for the black function, I'm gonna have to get two more than that. And so it's going to look, it's going to look like this. So let me see, I'm shifting it up by two. So for x less than negative three, it'll look like that. And for x greater than negative three, it is going to look like, it is going to look like that. | Graphing a shifted and stretched absolute value function.mp3 |
And so it's going to look, it's going to look like this. So let me see, I'm shifting it up by two. So for x less than negative three, it'll look like that. And for x greater than negative three, it is going to look like, it is going to look like that. And there you have it. This is the graph of y, or f of x, is equal to two times the absolute value of x plus three plus two. And you could have done it in different ways. | Graphing a shifted and stretched absolute value function.mp3 |
And for x greater than negative three, it is going to look like, it is going to look like that. And there you have it. This is the graph of y, or f of x, is equal to two times the absolute value of x plus three plus two. And you could have done it in different ways. You could have shifted up two first. Then you could have multiplied by a factor of two. And then you could have shifted, and then, so you could have moved up two first. | Graphing a shifted and stretched absolute value function.mp3 |
And you could have done it in different ways. You could have shifted up two first. Then you could have multiplied by a factor of two. And then you could have shifted, and then, so you could have moved up two first. Then you could have multiplied by a factor of two. Then you could have shifted left by three. Or you could have multiplied by a factor of two first, shifted up two, and then shifted over. | Graphing a shifted and stretched absolute value function.mp3 |
And then you could have shifted, and then, so you could have moved up two first. Then you could have multiplied by a factor of two. Then you could have shifted left by three. Or you could have multiplied by a factor of two first, shifted up two, and then shifted over. So there's multiple, there's three transformations going up here. This is an, this is a, let me color them all. So this right over here tells me, this over here says, hey, shift left. | Graphing a shifted and stretched absolute value function.mp3 |
Or you could have multiplied by a factor of two first, shifted up two, and then shifted over. So there's multiple, there's three transformations going up here. This is an, this is a, let me color them all. So this right over here tells me, this over here says, hey, shift left. Shift left by three. This told us stretch vertically by two, or essentially multiply the slope by two. Stretch vert by two. | Graphing a shifted and stretched absolute value function.mp3 |
The expression five times two to the t gives the number of leaves in a plant as a function of the number of weeks since it was planted. What does two represent in this expression? So pause this video and see if you can figure it out on your own. Alright, so let's look at the expression right over here. We could write it as defining a function. So we could say leaves as a function of time is equal to five times two to the t power. And so we could try this out a little bit. | Interpretting exponential expression.mp3 |
Alright, so let's look at the expression right over here. We could write it as defining a function. So we could say leaves as a function of time is equal to five times two to the t power. And so we could try this out a little bit. If we say, well what is L of zero? That would be t equals zero. That's when we're zero weeks after it was planted. | Interpretting exponential expression.mp3 |
And so we could try this out a little bit. If we say, well what is L of zero? That would be t equals zero. That's when we're zero weeks after it was planted. So this is right when it was planted. Well that's five times two to the zero, which is just, two to the zero is just one, so it's equal to five. And so when you see an exponential expression or an exponential function like this, that is why this number out here is often referred to as your initial value. | Interpretting exponential expression.mp3 |
That's when we're zero weeks after it was planted. So this is right when it was planted. Well that's five times two to the zero, which is just, two to the zero is just one, so it's equal to five. And so when you see an exponential expression or an exponential function like this, that is why this number out here is often referred to as your initial value. Initial value. And so let's explore this a little bit more. What is L of one? | Interpretting exponential expression.mp3 |
And so when you see an exponential expression or an exponential function like this, that is why this number out here is often referred to as your initial value. Initial value. And so let's explore this a little bit more. What is L of one? What happens after one week? Well that's gonna be five times two to the first power, or five times two. So going from when it was planted to the first week, we are multiplying by two, the number of leaves doubles. | Interpretting exponential expression.mp3 |
What is L of one? What happens after one week? Well that's gonna be five times two to the first power, or five times two. So going from when it was planted to the first week, we are multiplying by two, the number of leaves doubles. Well what happens after two weeks? The number of leaves after two weeks? Well that's gonna be five times two to the second power. | Interpretting exponential expression.mp3 |
So going from when it was planted to the first week, we are multiplying by two, the number of leaves doubles. Well what happens after two weeks? The number of leaves after two weeks? Well that's gonna be five times two to the second power. Well that's the number that you had in the first week times two. So it looks like every week we are doubling, we are multiplying by two. And that's why this number right over here, which is what the question is about, the two, this is often referred to as the common ratio. | Interpretting exponential expression.mp3 |
Well that's gonna be five times two to the second power. Well that's the number that you had in the first week times two. So it looks like every week we are doubling, we are multiplying by two. And that's why this number right over here, which is what the question is about, the two, this is often referred to as the common ratio. Common ratio. Because between any two successive weeks, the ratio between say week two and week one is two. Week two is double week one. | Interpretting exponential expression.mp3 |
And that's why this number right over here, which is what the question is about, the two, this is often referred to as the common ratio. Common ratio. Because between any two successive weeks, the ratio between say week two and week one is two. Week two is double week one. Week one is double week zero. So let's see which of these choices actually match up to that. There were initially two leaves in the plant. | Interpretting exponential expression.mp3 |
Week two is double week one. Week one is double week zero. So let's see which of these choices actually match up to that. There were initially two leaves in the plant. Well we know that there weren't two leaves in the plant, our initial value was five. So let me cross that one out. The number of leaves is multiplied by two each week. | Interpretting exponential expression.mp3 |
There were initially two leaves in the plant. Well we know that there weren't two leaves in the plant, our initial value was five. So let me cross that one out. The number of leaves is multiplied by two each week. Well that's exactly what we just described, so I like that choice. Let's look at the last one just for good measure. The plant was planted two weeks ago. | Interpretting exponential expression.mp3 |
The number of leaves is multiplied by two each week. Well that's exactly what we just described, so I like that choice. Let's look at the last one just for good measure. The plant was planted two weeks ago. Well no, they don't tell us anything about that. This is a general expression for T weeks after it was planted. So they're not saying when it was actually planted, so we could rule that out because we feel good about that second choice. | Interpretting exponential expression.mp3 |
And I encourage you to pause the video and think about it on your own. Well, there's a couple of ways to do this. One, you say, oh look, I'm multiplying two things that have the same base, so this is going to be that base, four, and then I add the exponents. Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. | Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3 |
Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. | Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3 |
Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. | Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3 |
And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a. | Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3 |