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# Square and Cube Root Function Families

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Extracting the Equation from a Graph

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 $y=a \sqrt{x-h}+k$ . The starting point is $(-6, 1)$ . Plugging this in for $h$ and $k$ , we have $y=a \sqrt{x+6}+1$ . Now, find $a$ , using the given point, $(-2, 5)$ . Let’s substitute it in for $x$ and $y$ and solve for $a$ .

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

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

#### Example B

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

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

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

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

#### Example C

Find the equation of the function below.

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

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

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

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

Intro Problem Revisit

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

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

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

### 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 $(-5, -3)$ and passes through $(4, -6)$ .

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

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

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

2. Substitute what you know into the general equation to solve for $a$ . From the graph, the starting point, or $(h, k)$ is $(4, -11)$ and $(13, 1)$ 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 $y=4 \sqrt{x-4}-11$ .

3. Substitute what you know into the general equation to solve for $a$ . From the graph, the starting point, or $(h, k)$ is $(-5, -3)$ and $(4, -6)$ 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 $y=- \sqrt{x+5}-3$ .

### Explore More

Write the equation for each function graphed below.

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