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# Square and Cube Root Function Families

## Identify and use h and k to determine equations

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

The equation is \begin{align*}y=2 \sqrt{x+6}+1\end{align*}.

#### Example B

Find the equation of the cubed root function where \begin{align*}h=-1\end{align*} and \begin{align*}k=-4\end{align*} and passes through \begin{align*}(-28, -3)\end{align*}.

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

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

#### Example C

Find the equation of the function below.

Solution: It looks like \begin{align*}(0, -4)\end{align*} is \begin{align*}(h, k)\end{align*}. Plug this in for \begin{align*}h\end{align*} and \begin{align*}k\end{align*} and then use the second point to find \begin{align*}a\end{align*}.

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

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

Intro Problem Revisit

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

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

### 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 \begin{align*}(-5, -3)\end{align*} and passes through \begin{align*}(4, -6)\end{align*}.

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

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

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

The equation of this function is \begin{align*}y=4 \sqrt{x-4}-11\end{align*}.

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

The equation of this function is \begin{align*}y=- \sqrt{x+5}-3\end{align*}.

### Explore More

Write the equation for each function graphed below.

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