# 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

In this concept, instead of graphing from the equation, we will now find the equation when we are given the graph.

Let's determine the equation of the graph below.

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*}.

\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*}.

Now, let's 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*}.

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*}.

\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*}.

Finally, let's 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 to find 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*}.

#### Example 2

Find the equation of the following function.

Substitute what you know into the general equation to solve for \begin{align*}a\end{align*}. From the final practice problem above, 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

Find the equation of the following function.

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.

### Notes/Highlights Having trouble? Report an issue.

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### 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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