Algebra 1 benchmark

Question

The equation below describes a function that models the height of a football thrown by a player. In the equation below, y represents the height of the football, in feet, and x represents the time the football is in the air, in seconds.

У = - 16x2 + 48x + 6

The maximum height reached by this football is ____

feet?

Answer 1

for y = -16x² + 48x + 6 → a = -16 and b = 48

the max height occurs at x = -b / 2a = -48 / {(2)(-16)} = 48 / 32 = 1.5

Use 1.5 for x in the equation to determine the height of the ball at its peak.

y = -16(1.5)² + 48(1.5) + 6 = -36 + 72 +6 = 42

Think about this: While the ball is on an upward trajectory, the slope of y is positive. When it is on a downward trajectory, the slope is negative. When the slope is zero, the ball is at its peak. From calculus, the derivative of a function is its slope. The derivative set to zero is

y' = -32x + 48 =0 or x 48 / 32 = 1.5 (same as above)

Answer 2

Hi Jess T.

42 ft

y = - 16x² + 48x + 6

First you have as quadratic in the form of

y= ax² + bx + c

This is a parabola

a= -16 and less than zero your parabola opens downward, so the x or h component of the vertex of your downward parabola will will represent the maximum height. 48 is the coefficient of b.

y= ax² + bx + c

x or h component of your parabola is -b/2a

-48/2(-16) = -48/-32 = 48/32 = 3/2

The y component of your vertex can be found by plugging x into your original quadratic

y = -16(3/2)² + 48(3/2) + 6

y = -16(9/4) + 72 + 6

y = -4(9) + 72 + 6

y = -36 + 72 + 6

y = 36 + 6

y = 42 ft

You can graph your function at Desmos.com to confirm the coordinates of the vertex. Hope this helpts