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# 7.6: Extracting the Equation from a Graph

Difficulty Level: At Grade Created by: CK-12

The graph of a cubic function starts at the point (2, 2). It passes through the point (10, -2). What is the equation of the function?

### Guidance

This concept is the opposite of the previous two. Instead of graphing from the equation, we will now find the equation, given the graph.

#### Example A

Determine the equation of the graph below.

Solution: From the previous two concepts, we know this is a square root function, so the general form is y=axh+k$y=a \sqrt{x-h}+k$ . The starting point is (6,1)$(-6, 1)$ . Plugging this in for h$h$ and k$k$ , we have y=ax+6+1$y=a \sqrt{x+6}+1$ . Now, find a$a$ , using the given point, (2,5)$(-2, 5)$ . Let’s substitute it in for x$x$ and y$y$ and solve for a$a$ .

5&=a \sqrt{-2+6}+1 \\
4&=a \sqrt{4} \\
4&=2a \\
2&=a

The equation is y=2x+6+1$y=2 \sqrt{x+6}+1$ .

#### Example B

Find the equation of the cubed root function where h=1$h=-1$ and k=4$k=-4$ and passes through (28,3)$(-28, -3)$ .

Solution: First, plug in what we know to the general equation; y=xh3+ky=ax+134$y= \sqrt[3]{x-h}+k \Rightarrow y=a \sqrt[3]{x+1}-4$ . Now, substitute x=28$x=-28$ and y=3$y=-3$ and solve for a$a$ .

-3&=a \sqrt[3]{-28+1}-4 \\
1&=-3a \\
- \frac{1}{3}&=a

The equation of the function is y=13x+134$y=- \frac{1}{3} \sqrt[3]{x+1}-4$ .

#### Example C

Find the equation of the function below.

Solution: It looks like (0,4)$(0, -4)$ is (h,k)$(h, k)$ . Plug this in for h$h$ and k$k$ and then use the second point to find a$a$ .

-6&=a \sqrt[3]{1-0}-4 \\
-2&=a \sqrt[3]{1} \\
-2&=a

The equation of this function is y=2x34$y=-2 \sqrt[3]{x}-4$ .

When finding the equation of a cubed root function, you may assume that one of the given points is (h,k)$(h, k)$ . Whichever point is on the “bend” is (h,k)$(h, k)$ for the purposes of this text.

Intro Problem Revisit

First, plug in what we know to the general equation; y=xh3+ky=ax23+2$y= \sqrt[3]{x-h}+k \Rightarrow y=a \sqrt[3]{x - 2} + 2$ . Now, substitute x=10$x=10$ and y=2$y=-2$ and solve for a$a$ .

-2&=a \sqrt[3]{10-2}+2 \\
-2&=a\sqrt[3]{8}+2\\
-2&=2a +2\\
-4&=2a\\
a = -2

The equation of the function is y=2x23+2$y=-2 \sqrt[3]{x-2}+2$ .

### Guided Practice

Find the equation of the functions below.

1.

2.

3. Find the equation of a square root equation with a starting point of (5,3)$(-5, -3)$ and passes through (4,6)$(4, -6)$ .

1. Substitute what you know into the general equation to solve for a$a$ . From Example C, you may assume that (5,8)$(5, 8)$ is (h,k)$(h, k)$ and (3,7)$(-3, 7)$ is (x,y)$(x, y)$ .

y&=a \sqrt[3]{x-5}+8 \\
7&=a \sqrt[3]{-3-5}+8 \\
-1&=-2a \\
\frac{1}{2}&=a

The equation of this function is y=12x53+8$y= \frac{1}{2} \sqrt[3]{x-5}+8$ .

2. Substitute what you know into the general equation to solve for a$a$ . From the graph, the starting point, or (h,k)$(h, k)$ is (4,11)$(4, -11)$ and (13,1)$(13, 1)$ are a point on the graph.

y&=a \sqrt{x-4}-11 \\
1&=a \sqrt{13-4}-11 \\
12&=3a \\
4&=a

The equation of this function is y=4x411$y=4 \sqrt{x-4}-11$ .

3. Substitute what you know into the general equation to solve for a$a$ . From the graph, the starting point, or (h,k)$(h, k)$ is (5,3)$(-5, -3)$ and (4,6)$(4, -6)$ are a point on the graph.

y&=a \sqrt{x+5}-3 \\
-6&=a \sqrt{4+5}-3 \\
-3&=3a \\
-1&=a

The equation of this function is y=x+53$y=- \sqrt{x+5}-3$ .

### Explore More

Write the equation for each function graphed below.

1. Write the equation of a square root function with starting point (6,3)$(-6, -3)$ passing through (10,15)$(10, -15)$ .
2. Write the equation of a cube root function with (h,k)=(2,7)$(h, k) = (2, 7)$ passing through (10,11)$(10, 11)$ .
3. Write the equation of a square root function with starting point (1,6)$(-1, 6)$ passing through (3,16)$(3, 16)$ .
4. Write the equation of a cubed root function with (h,k)=(1,6)$(h, k)=(-1, 6)$ passing through (7,16)$(7, 16)$ .
5. Write the equation of a cubed root function with (h,k)=(7,16)$(h, k)=(7, 16)$ passing through (1,6)$(-1, 6)$ .
6. How do the two equations above differ? How are they the same?

Mar 12, 2013

Jun 04, 2015

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