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y = 2 x 2 4 x + 1

y : ( 0 , 1 ) ; x : ( 1.7 , 0 ) , ( 0.3 , 0 ) ;
axis: x = 1 ; vertex : ( 1 , −1 )
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, -1) and the intercepts (1.7, 0), (0.3, 0) and (0, 1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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y = 2 x 2 4 x + 2

y : ( 0 , 2 ) x : ( 1 , 0 ) ;
axis: x = 1 ; vertex: ( 1 , 0 )
 This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 0) and the intercept (0, 2). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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y = −4 x 2 6 x 2

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y = x 2 4 x + 2

y : ( 0 , 2 ) x : ( −4.4 , 0 ) , ( 0.4 , 0 ) ;
axis: x = −2 ; vertex: ( −2 , 6 )
This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (-2, 6) and the intercepts (-4.4, 0), (0.4, 0) and (0, 2). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2.

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y = 5 x 2 10 x + 8

y : ( 0 , 8 ) ; x : none ;
axis: x = 1 ; vertex : ( 1 , 3 )
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 3) and the intercept(0, 8). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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y = −16 x 2 + 24 x 9

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y = 3 x 2 + 18 x + 20

y : ( 0 , 20 ) x : ( −4.5 , 0 ) , ( −1.5 , 0 ) ;
axis: x = −3 ; vertex: ( −3 , −7 )
This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (-3, -7) and the intercepts (-4.5, 0) and (-1.5, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -3.

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y = −2 x 2 + 8 x 10

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Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value.

y = 2 x 2 + x 1

The minimum value is 9 8 when x = 1 4 .

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y = −4 x 2 + 12 x 5

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y = x 2 6 x + 15

The minimum value is 6 when x = 3 .

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y = −9 x 2 + 16

The maximum value is 16 when x = 0 .

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In the following exercises, solve. Round answers to the nearest tenth.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation h = −16 t 2 + 168 t + 45 to find how long it will take the arrow to reach its maximum height, and then find the maximum height.

In 5.3 sec the arrow will reach maximum height of 486 ft.

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A stone is thrown vertically upward from a platform that is 20 feet high at a rate of 160 ft/sec. Use the quadratic equation h = −16 t 2 + 160 t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.

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A computer store owner estimates that by charging x dollars each for a certain computer, he can sell 40 x computers each week. The quadratic equation R = x 2 + 40 x is used to find the revenue, R , received when the selling price of a computer is x . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

20 computers will give the maximum of $400 in receipts.

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A retailer who sells backpacks estimates that, by selling them for x dollars each, he will be able to sell 100 x backpacks a month. The quadratic equation R = x 2 + 100 x is used to find the R received when the selling price of a backpack is x . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

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A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation A = x ( 240 2 x ) gives the area of the corral, A , for the length, x , of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

The length of the side along the river of the corral is 120 feet and the maximum area is 7,200 sq ft.

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A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation A = x ( 100 2 x ) gives the area, A , of the dog run for the length, x , of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

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Everyday math

In the previous set of exercises, you worked with the quadratic equation R = x 2 + 40 x that modeled the revenue received from selling computers at a price of x dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
Graph the equation R = x 2 + 40 x . Find the values of the x -intercepts.


  1. This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 60. The y-axis of the plane runs from -50 to 500. The parabola has a vertex at (20, 400) and also goes through the points (0, 0) and (40, 0).
  2. ( 0 , 0 ) , ( 40 , 0 )
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Practice Key Terms 6

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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