A quadratic function is shown below. h(t) = –16t^2 + 128t

Question

A quadratic function is shown below.

h(t) = –16t² + 128t

The function models the height in feet, h(t), an object is above the ground in terms of elapsed time in seconds, t, from when it is projected directly upward. The object has an initial velocity of 128 feet/second.

 

Part A

Graph h(t) on the coordinate grid. Label and scale axes appropriately.

  

Part B

Identify the x-intercept(s) or zero(s) of the function and explain the significance in terms of the problem situation.

   

Part C

Identify the y-intercept of the function and explain the significance in terms of the problem situation.

 

 

Part D

Identify the vertex of the function, including whether it is a minimum or maximum, and explain the significance in terms of the problem situation.

 

Part E

Identify the axis of symmetry of the function and explain the significance in terms of the problem situation.

 

Answer 1

A. Desmos graph

B. X intercepts are the two places where parabola intersects line y=0. This is where the object is at ground level, the intercepts (0,0) and (8,0) where x is time in seconds. One is where the object starts its movement and the other is when it stops.

C. Y intercept is the starting point for the object where x(time)=0.

D. The vertex is at the peak of the graph. To find it use -b/2a to get the x coordinate, substitute it into the problem to the y coordinate. (4,256)

-128/2(-16) = -128/-32 = 4 -16(4)^2 + 128(4) = 256

This represents a maximum because the parabola has a peak. If a is negative it is a maximum, if it positive there is a minimum.

E. The axis of symmetry is the x value of the vertex as an equation, so it is x=4. The axis of symmetry bisects the graph making every coordinate on the left have a mirrored point on the right.