# 7.6: Extracting the Equation from a Graph

**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?

### Extracting the Equation from a Graph

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

#### Solve the following problems

Determine the equation of the graph below.

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*}

\begin{align*}5&=a \sqrt{-2+6}+1 \\
4&=a \sqrt{4} \\
4&=2a \\
2&=a\end{align*}

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

Find the equation of the cubed root function where \begin{align*}h=-1\end{align*}

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*}

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

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

Find the equation of the function below.

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*}.

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

### Examples

#### Example 1

Earlier, you were asked what is the equation of the function.

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*}.

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

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

Find the equation of the functions below.

#### Example 2

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*}.

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

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

#### Example 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*}(4, -11)\end{align*} and \begin{align*}(13, 1)\end{align*} are a point on the graph.

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

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

#### Example 4

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*}.

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.

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

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

### Review

Write the equation for each function graphed below.

- 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*}.
- 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*}.
- 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*}.
- 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*}.
- 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*}.
- How do the two equations above differ? How are they the same?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.6.

### Image Attributions

Here you'll look at the graph of a square root or cubed root function and determine the equation.

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