few_shots: Example 1:
Problem/activity:
Task
Materials
One 10-frame per pair of students (see example below)
[IMG]
Counters, 10 per pair of students
Decks of number cards 0-5 (with 4 of each number) or two 0-5 dice
Setup
This activity is done in pairs. The students sit side by side with 10-frame and deck of cards or dice. If they have cards, the cards are shuffled and placed face down in front of them.
Action
Player 1 flips a card or rolls the die. They place that many counters on the 10-frame. Player 2 flips a card or rolls the die. They place that many more counters on the 10-frame next to the counters already on the ten frame. The students count the total together.
A recording sheet can be added when the students have been introduced to equations and number sentences. For those students who can handle higher numbers, a deck of 0-10 cards and two 10-frames can be used.
This game can also be played alone.


Standard options: 
A. Represent addition and subtraction with objects, fingers, mental images, drawings (Drawings need not show details, but should show the mathematics in the problem. (This applies wherever drawings are mentioned in the Standards.)), sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
B. For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
C. Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1).
D. Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
E. Fluently add and subtract within 5.


Your response:
{
  "explanation": "This task involves representing addition with objects such as number cards and dice.",
  "answer": "A"
}

Example 2:
Problem/activity:
For each pair of numbers, determine if the values are equivalent. Explain your reasoning for each one.
a.
$$-0.25$$
and
$$\frac{1}{-4}$$
b.
$$-\left ( \frac{1}{4} \right )$$
and
$$\frac{-1}{-4}$$
c.
$$\frac{-1}{4}$$
and
$$-\frac{1}{4}$$


Standard options: 
A. Solve real-world and mathematical problems involving the four operations with rational numbers. (Computations with rational numbers extend the rules for manipulating fractions to complex fractions.)
B. Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.
C. Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.


Your response:
{
  "explanation": "This problem involves understanding fractions and division to divide rational numbers, such as 1, -4, 4, and -1. This problem also involves an understanding of multiplication: if p and q are integers, then -(p/q) = (-p)/q = p/(-q).",
  "answer": "B"
}

Example 3:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Standard options: 
A. Prove the Pythagorean identity sin^(2)(θ) + cos^(2)(θ) = 1 and use it to find sin(θ), cos(θ),or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.
B. Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.


Your response:
{
  "explanation": "An initial step when solving for x in this problem requires the use of the addition formula for cosine on the left side of the equation. Solving this problem does not require use of the Pythagorean identity.",
  "answer": "B"
}

Now, assign the following problem or activity to one or more relevant standards it aligns with.
shot_info: im_task_001477

K

fl_problem_001097

7

fl_problem_001894

HS_0
options: Graph the solutions to a linear inequality in two variables as a half plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes.

Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).
problem_activity: Activity
By now, students recognize that when a quadratic equation is in the form of
\(\text{expression} = 0\)
and the expression is in factored form, the equation can be solved using the zero product property. In this activity, they encounter equations in which one side of the equal sign is not 0. To make one side equal 0 requires rearrangement. For example, to solve
\(x(x+6)=8\)
, the equation needs to be rearranged to
\(x(x+6)-8=0\)
. Yet because the expression on the left is no longer in factored form, the zero product property won’t help after all and another strategy is needed.
Students recall that to solve a quadratic equation in the form of
\(\text{expression} = 0\)
is essentially to find the zeros of a quadratic function defined by that expression, and that the zeros of a function correspond to the horizontal intercepts of its graph. In the case of
\(x(x+6)-8=0\)
, the function whose zeros we want to find is defined by
\(x(x+6)-8\)
. Graphing
\(y=x(x+6)-8\)
and examining the
\(x\)
-intercepts of the graph allow us to see the number of solutions and what they are.
Launch
Arrange students in groups of 2 and provide access to graphing technology. Give students a moment to think quietly about the first question and then ask them to briefly discuss their response with their partner before continuing with the rest of the activity.
Action and Expression: Provide Access for Physical Action.
Support effective and efficient use of tools and assistive technologies. To use graphing technology, some students may benefit from a demonstration or access to step-by-step instructions.
Supports accessibility for: Organization; Memory; Attention
Student Facing
Han is solving three equations by graphing.
\((x-5)(x-3)=0\)
\(\displaystyle (x-5)(x-3)=\text-1\)
\(\displaystyle (x-5)(x-3)=\text-4\)
To solve the first equation,
\((x-5)(x-3)=0\)
, he graphed
\(y=(x-5)(x-3)\)
and then looked for the
\(x\)
-intercepts of the graph.
Explain why the
\(x\)
-intercepts can be used to solve
\((x-5)(x-3)=0\)
.
What are the solutions?
To solve the second equation, Han rewrote it as
\((x-5)(x-3)+1=0\)
. He then graphed
\(y=(x-5)(x-3)+1\)
.
Use graphing technology to graph
\(y=(x-5)(x-3)+1\)
. Then, use the graph to solve the equation. Be prepared to explain how you use the graph for solving.
Solve the third equation using Han’s strategy.
Think about the strategy you used and the solutions you found.
Why might it be helpful to rearrange each equation to equal 0 on one side and then graph the expression on the non-zero side?
How many solutions does each of the three equations have?
Student Response
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Student Facing
Are you ready for more?
The equations
\((x-3)(x-5)=\text-1\)
,
\((x-3)(x-5)=0\)
, and
\((x-3)(x-5)=3\)
all have whole-number solutions.
Use graphing technology to graph each of the following pairs of equations on the same coordinate plane. Analyze the graphs and explain how each pair helps to solve the related equation.
\(y=(x-3)(x-5)\)
and
\(y=\text-1\)
\(y=(x-3)(x-5)\)
and
\(y=0\)
\(y=(x-3)(x-5)\)
and
\(y=3\)
Use the graphs to help you find a few other equations of the form
\((x-3)(x-5)=z\)
that have whole-number solutions.
Find a pattern in the values of
\(z\)
that give whole-number solutions.
Without solving, determine if
\((x-5)(x-3)=120\)
and
\((x-5)(x-3)=399\)
have whole-number solutions. Explain your reasoning.
Student Response
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click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
If students enter the equation
\((x-5)(x-3)=0\)
into their graphing technology, they may see an error message, or they may see vertical lines. The lines will intersect the
\(x\)
-axis at the solutions, but they are clearly not graphs of a quadratic function. Emphasize that we want to graph the function defined by
\(y=(x-5)(x-3)\)
and use its
\(x\)
-intercepts to find the solution to the related equation.
All
the points on the two vertical lines do represent solutions to the equation (because the points along each vertical line satisfy the equation regardless of the value chosen for
\(y\)
), but understanding this is beyond the expectations for students in this course.
Activity Synthesis
Invite students to share their responses, graphs, and explanations on how they used the graphs to solve the equations. Discuss questions such as:
“
Are the original equation
\((x-5)(x-3)=\text-1\)
and the rewritten one
\((x-5)(x-3)+1=0\)
equivalent?” (Yes, each pair of equations are equivalent. In that example, 1 is added to both sides of the original equation.)
"Why might it be helpful to rearrange the equation so that one side is an expression and the other side is 0?" (It allows us to find the zeros of the function defined by that expression. The zeros correspond to the
\(x\)
-intercepts of the graph.)
“What equation would you graph to solve this equation:
\((x-4)(x-6)=15\)
?” (
\(y=(x-4)(x-6)-15\)
) “What about
\((x+3)^2-1=5\)
?" (
\(y=(x+3)^2-6\)
)
Make sure students understand that some quadratic functions have two zeros, some have one zero, and some have no zeros, so their respective graphs will have two, one, or no horizontal intercepts, respectively.
Likewise, some quadratic equations have two solutions, some have one solution, and some have no real solutions. (Because students won’t know about numbers that aren’t real until a future course, for now it is sufficient to say “no solutions.”)


correct_option_index: 0

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