In one sentence, describe what the following article is about: There are several important considerations here: Do you want to hold it during school term or during the vacation time? During school term limits the options to weekend nights, while during school vacation, some of the friends might be away. Are there any big events on that might stop guests from coming? For example, big sports meet-ups, local drama plays being staged, try-outs for teams, etc. Are there known issues about some friends being tied up doing something on any particular dates? Especially if they're BFFs, don't plan the party if they're not going to be able to make it. Is it near or around exam time? If so, do not hold the party as it will distract the girls and many of them probably won't be allowed to come anyway. Is the weather going to be appropriate for the sleepover? Although the girls sleep indoors, this could impact outdoor activities or outings to things. If you're a parent planning the party, ask your daughter if she knows of anything that might prevent guests from being available on a specific date. If not, and you've considered the above possibilities, you're good to go. A late afternoon start is fairly standard, such as around 4pm, 5pm, etc. This allows time for the girls to settle in or for you take them somewhere, such as to see a movie, to go bowling or to have a pizza meal out. Obviously, the time will depend on whether or not you've planned an external activity. Be sure to seek the permission of parents if taking the girls out of your home. They may prefer to help with transportation arrangements. It's nice to let the parents have a little sleep in or take their time to come and collect their daughter the next morning and the girls usually love staying in one another's company a little longer. A 10am, 11am or noon pick-up time usually works best. It does mean that you have to include breakfast.
Summary: Decide when it's best to hold the slumber party. Consider what time you'd like the girls to come to the sleepover. Consider how long you would like the girls to stay the following morning.

Problem: Article: A graph makes it easier to follow the problem and check whether the answer makes sense. Sketch the function on a piece of graph paper, using a graphing calculator as a reference if necessary. Sketch the tangent line going through the given point. (Remember, the tangent line runs through that point and has the same slope as the graph at that point.) Example 1: Sketch the graph of the parabola f(x)=0.5x2+3x1{\displaystyle f(x)=0.5x^{2}+3x-1}. Draw the tangent going through point (-6, -1).You don't know the tangent's equation yet, but you can already tell that its slope is negative, and that its y-intercept is negative (well below the parabola vertex with y value -5.5). If your final answer doesn't match these details, you'll know to check your work for mistakes. For function f(x), the first derivative f'(x) represents the equation for the slope of the tangent line at any point on f(x). There are many ways to take derivatives. Here's a simple example using the power rule: Example 1 (cont.): The graph is described by the function f(x)=0.5x2+3x1{\displaystyle f(x)=0.5x^{2}+3x-1}.Recall the power rule when taking derivatives: ddxxn=nxn1{\displaystyle {
rac {d}{dx}}x^{n}=nx^{n-1}}.The function's first derivative = f'(x) = (2)(0.5)x + 3 - 0.f'(x) = x + 3. Plug any value a for x into this equation, and the result will be the slope of the line tangent to f(x) at the point were x = a. Read the problem to discover the coordinates of the point for which you're finding the tangent line. Enter the x-coordinate of this point into f'(x). The output is the slope of the tangent line at this point. Example 1 (cont.): The point mentioned in the problem is (-6, -1). Use the x-coordinate -6 as the input for f'(x):f'(-6) = -6 + 3 = -3The slope of the tangent line is -3. The point-slope form of a linear equation is yy1=m(xx1){\displaystyle y-y_{1}=m(x-x_{1})}, where m is the slope and (x1,y1){\displaystyle (x_{1},y_{1})} is a point on the line. You now have all the information you need to write the tangent line's equation in this form. Example 1 (cont.): yy1=m(xx1){\displaystyle y-y_{1}=m(x-x_{1})}The slope of the line is -3, so yy1=3(xx1){\displaystyle y-y_{1}=-3(x-x_{1})}The tangent line passes through (-6, -1), so the final equation is y(1)=3(x(6)){\displaystyle y-(-1)=-3(x-(-6))}Simplify to y+1=3x18{\displaystyle y+1=-3x-18}y=3x19{\displaystyle y=-3x-19} If you have a graphing calculator, graph the original function and the tangent line to check that you have the correct answer. If working on paper, refer to your earlier graph to make sure there are no glaring mistakes in your answer. Example 1 (cont.): The initial sketch showed that the slope of the tangent line was negative, and the y-intercept was well below -5.5. The tangent line equation we found is y = -3x - 19 in slope-intercept form, meaning -3 is the slope and -19 is the y-intercept. Both of these attributes match the initial predictions. Here's a run-through of the whole process again. This time, the goal is to find the line tangent to f(x)=x3+2x2+5x+1{\displaystyle f(x)=x^{3}+2x^{2}+5x+1} at x = 2: Using the power rule, the first derivative f(x)=3x2+4x+5{\displaystyle f'(x)=3x^{2}+4x+5}. This function will tell us the slope of the tangent. Since x = 2, find f(2)=3(2)2+4(2)+5=25{\displaystyle f'(2)=3(2)^{2}+4(2)+5=25}. This is the slope at x = 2. Notice we do not have a point this time, only an x-coordinate. To find the y-coordinate, plug x = 2 into the initial function: f(2)=23+2(2)2+5(2)+1=27{\displaystyle f(2)=2^{3}+2(2)^{2}+5(2)+1=27}. The point is (2,27). Write the tangent line equation in point-slope form: yy1=m(xx1){\displaystyle y-y_{1}=m(x-x_{1})}y27=25(x2){\displaystyle y-27=25(x-2)}If required, simplify to y = 25x - 23.
Summary:
Sketch the function and tangent line (recommended). Take the first derivative to find the equation for the slope of the tangent line. Enter the x value of the point you're investigating. Write the tangent line equation in point-slope form. Confirm the equation on your graph. Try a more difficult problem.