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Square and Cube Root Function Families

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Extracting the Equation from a Graph
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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) .

Answers

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 .

Practice

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?

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