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## Identify vertices and up and down parabolas

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On reduced-gravity flights, airplanes fly in large parabolic arcs. Determine the quadratic equation of best fit for the data set below that represents the arc in which the airplane flies.

x\begin{align*}x \end{align*} (Time in Sec.) 0 20 33 45
y\begin{align*}y \end{align*} (Altitude in 1000 feet) 24 29 33 29

When finding the equation of a parabola, you can use any of the three forms. If you are given the vertex and any other point, you only need two points to find the equation. However, if you are not given the vertex you must have at least three points to find the equation of a parabola.

Let's find the equation of the parabola with vertex (-1, -4) and that passes through (2, 8).

Use vertex form and substitute -1 for h\begin{align*}h\end{align*} and -4 for k\begin{align*}k\end{align*}.

yy=a(x(1))24=a(x+1)24\begin{align*}y &=a(x-(-1))^2-4\\ y &=a(x+1)^2-4\end{align*}

Now, take the second point and plug it for x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} and solve for a\begin{align*}a\end{align*}.

81213=a(2+4)24=36a=a\begin{align*}8 &=a(2+4)^2-4\\ 12 &=36a\\ \frac{1}{3} &=a\end{align*}

The equation is y=13(x+1)24\begin{align*}y=\frac{1}{3}(x+1)^2-4\end{align*}.

We can fit a set of data to a quadratic equation. In this concept, we will be using quadratic regression and a TI-83/84.

Now, let's determine the quadratic equation of best fit for the data set below.

x\begin{align*}x\end{align*} 0 4 7 12 17
y\begin{align*}y\end{align*} 7 9 10 8 3

We need to enter the x\begin{align*}x-\end{align*}coordinates as a list of data and the y\begin{align*}y-\end{align*}coordinates as another list.

Step 1: Press STAT.

Step 2: In EDIT, select 1:Edit…. Press ENTER.

Step 3: The List table appears. If there are any current lists, you will need to clear them. To do this, arrow up to L1 so that it is highlighted (black). Press CLEAR, then ENTER. Repeat with L2, if necessary.

Step 4: Now, enter the data into the lists. Enter all the entries into L1 (x)\begin{align*}(x)\end{align*} first and press enter between each entry. Then, repeat with L2 and y\begin{align*}y\end{align*}.

Step 5: Press 2nd\begin{align*}2^{nd}\end{align*} MODE (QUIT).

Now that we have everything in the lists, we can use quadratic regression to determine the equation of best fit.

Step 6: Press STAT and then arrow over to the CALC menu.

Step 7: Select 5:QuadReg. Press ENTER.

Step 8: You will be taken back to the main screen. Type (L1,L2) and press ENTER. L1 is 2nd\begin{align*}2^{nd}\end{align*} 1, L2 is 2nd\begin{align*}2^{nd}\end{align*} 2.

Step 9: The following screen appears. The equation of best fit is y=0.64x2+0.86x+6.90\begin{align*}y=-0.64x^2+0.86x+6.90\end{align*}.

This technique can be applied to real-life problems, and you can also use it to find the equation of any parabola, given three points.

Let's find the equation of the parabola that passes through (1, 11), (2, 20), (-3, 75).

You can use the same steps from above to find the equation of the parabola. Doing this, you should get the equation is y=5x26x+12\begin{align*}y=5x^2-6x+12\end{align*}.

This problem can also be done by solving three equations, with three unknowns. If we plug in (x,y)\begin{align*}(x, y)\end{align*} to y=ax2+bx+c\begin{align*}y=ax^2+bx+c\end{align*}, we would get:

112075=a+b+c=4a+2b+c=9a3b+c\begin{align*}11 &=a+b+c\\ 20 &=4a+2b+c\\ 75 &=9a-3b+c\end{align*}

Use linear combinations to solve this system of equations. This problem will be finished in the Review section (#13).

### Examples

#### Example 1

Earlier, you were asked to determine the quadratic equation of best fit for the data given.

Use your calculator to find the quadratic equation of best fit for the given table.

x\begin{align*}x \end{align*} (Time in Sec.) 0 20 33 45
y\begin{align*}y \end{align*} (Altitude in 1000 feet) 24 29 33 29

y=0.0081x2+0.495x+23.6987\begin{align*}y = -0.0081x^2 + 0.495x + 23.6987\end{align*} is the quadratic equation of best fit for the data.

#### Example 2

Find the equation of the parabola with x\begin{align*}x-\end{align*}intercepts (4, 0) and (-5, 0) that passes through (-3, 8).

Because we are given the intercepts, use intercept form to find the equation.

y=a(x4)(x+5)\begin{align*}y =a(x-4)(x+5)\end{align*}

Plug in (-3,8) and solve for a\begin{align*}a\end{align*}

8847=a(34)(3+5)=14a=a\begin{align*}8 &=a(-3-4)(-3+5)\\ 8 &=-14a\\ -\frac{4}{7} &=a\end{align*}

The equation of the parabola is y=47(x4)(x+5)\begin{align*}y=- \frac{4}{7}(x-4)(x+5)\end{align*}.

#### Example 3

A study compared the speed, x\begin{align*}x\end{align*} (in miles per hour), and the average fuel economy, y\begin{align*}y\end{align*} (in miles per gallon) of a sports car. Here are the results.

speed 30 40 50 55 60 65 70 80
fuel economy 11.9 16.1 21.1 22.2 25.0 26.1 25.5 23.2

Plot the scatterplot and use your calculator to find the equation of best fit.

Plotting the points, we have:

Using the steps from above, the quadratic regression equation is y=0.009x2+1.24x18.23\begin{align*}y=-0.009x^2+1.24x-18.23\end{align*}.

### Review

Find the equation of the parabola given the following points. No decimal answers.

1. vertex: (-1, 1) point: (1, -7)
2. x\begin{align*}x-\end{align*}intercepts: -2, 2 point: (4, 3)
3. vertex: (9, -4) point: (5, 12)
4. x\begin{align*}x-\end{align*}intercepts: 8, -5 point: (3, 20)
5. x\begin{align*}x-\end{align*}intercepts: -9, -7 point: (-3, 36)
6. vertex: (6, 10) point: (2, -38)
7. vertex: (-4, -15) point: (-10, 1)
8. vertex: (0, 2) point: (-4, -12)
9. x\begin{align*}x-\end{align*}intercepts: 3, 16 point: (7, 24)

Use a graphing calculator to find the quadratic equation (in standard form) that passes through the given three points. No decimal answers.

1. (-4, -51), (-1, -18), (4, -43)
2. (-5, 131), (-1, -5), (3, 51)
3. (-2, 9), (2, 13), (6, 41)
4. Challenge Finish computing the equation of the parabola that passes through (1, 11), (2, 20), (-3, 75) using linear combinations.

For the quadratic modeling questions below, use a graphing calculator. Round any decimal answers to the nearest hundredth.

1. The surface of a speed bump is shaped like a parabola. Write a quadratic model for the surface of the speed bump shown.
2. Physics and Photography Connection Your physics teacher gives you a project to analyze parabolic motion. You know that when a person throws a football, the path is a parabola. Using your camera, you take an long exposure picture of a friend throwing a football. A sketch of the picture is below. You put the path of the football over a grid, with the x\begin{align*}x-\end{align*}axis as the horizontal distance and the y\begin{align*}y-\end{align*}axis as the height, both in 3 feet increments. The release point, or shoulder height, of your friend is 5 ft, 3 in and you estimate that the maximum height is 23 feet. Find the equation of the parabola.
3. An independent study was done linking advertising to the purchase of an object. 400 households were used in the survey and the commercial exposure was over a one week period. See the data set below.
# of times commercial was shown, x\begin{align*}x\end{align*} 1 7 14 21 28 35 42 49
# of households bought item, y\begin{align*}y\end{align*} 2 25 96 138 88 37 8 6

a) Find the quadratic equation of best fit.

b) Why do you think the amount of homes that purchased the item went down after more exposure to the commercial?

To see the Review answers, open this PDF file and look for section 5.18.

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