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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?

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

\begin{align*}5&=a \sqrt{-2+6}+1 \\ 4&=a \sqrt{4} \\ 4&=2a \\ 2&=a\end{align*}5442=a2+6+1=a4=2a=a

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

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

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*}y=xh3+ky=ax+134. 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*}.

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.

  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?

Answers for Review Problems

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

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Difficulty Level:
At Grade
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Date Created:
Mar 12, 2013
Last Modified:
Jun 07, 2016

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