Question

complete problems 1 - 5. you must draw a sketch with every question that accurately represents the problem.

1. bobby hits a golf ball. the following function models the height, ( f(x) ), in feet, of an object ( x ) seconds after it is hit in the air: ( f(x)=-16 x^{2}+96 x ).

a. what time is the golf ball highest in the air?
b. how high is it at it highest point?

1. jason heyward hits a baseball. the following function models the height, ( h(t) ), in feet, of an object ( t ) seconds after it is hit in the air: ( f(t)=-16 t^{2}+64 t + 3 ).

a. what is the highest the baseball will go?
b. is the ball going up or down after 2.5 seconds?

1. a shoe is thrown off a cliff. the following function models the height, ( f(x) ), in feet, of an object ( x ) seconds after it is in the air: ( f(x)=-16 x^{2}+24 x + 150 ).

a. is the shoe going up, down, or at its highest point after exactly 3 seconds?
b. how high is it after exactly 3 seconds?

1. bozo the clown is shot out of a human canon at the circus. the following function models the height, ( h(t) ), in feet, of an object ( t ) seconds after it is in the air: ( f(t)=-16 t^{2}+32 t + 5 ).

a. is bozo going up or down after exactly 2 seconds?
b. how high off the ground is he after exactly 2 seconds?

Explanation:

Step1: Analyze the function form

The given functions are of the form \(y = ax^{2}+bx + c\), which is a quadratic function. For a quadratic function \(y = ax^{2}+bx + c\), the vertex of the parabola (which gives the maximum height for \(a<0\)) has its \(x\)-coordinate (in the case of time - related problems) given by \(x=-\frac{b}{2a}\).

Problem 1:

• Part a

For the function \(f(x)=-16x^{2}+96x\), where \(a=-16\) and \(b = 96\)

$$x=-\frac{96}{2\times(-16)}=\frac{-96}{-32}=3$$

• Part b

Substitute \(x = 3\) into \(f(x)=-16x^{2}+96x\)

$$f(3)=-16\times3^{2}+96\times3=-16\times9 + 288=-144+288 = 144$$

Problem 2:

• Part a

For the function \(f(t)=-16t^{2}+64t + 3\), where \(a=-16\) and \(b = 64\)

$$t=-\frac{64}{2\times(-16)}=\frac{-64}{-32}=2$$

Substitute \(t = 2\) into \(f(t)\)

$$f(2)=-16\times2^{2}+64\times2+3=-16\times4+128 + 3=-64 + 128+3=67$$

• Part b

The axis of symmetry is \(t = 2\). Since the parabola \(y=-16t^{2}+64t + 3\) opens downwards (\(a=-16<0\)), and \(2.5>2\), the ball is going down after \(t = 2.5\) seconds.

Problem 3:

• Part a

For the function \(f(x)=-16x^{2}+24x + 150\), where \(a=-16\) and \(b = 24\)
The axis of symmetry is \(x=-\frac{24}{2\times(-16)}=\frac{-24}{-32}=0.75\). Since \(3>0.75\) and \(a=-16<0\) (parabola opens downwards), the shoe is going down at \(x = 3\) seconds.

• Part b

Substitute \(x = 3\) into \(f(x)=-16x^{2}+24x + 150\)

$$f(3)=-16\times3^{2}+24\times3+150=-16\times9+72 + 150=-144+72+150=78$$

Problem 4:

• Part a

For the function \(f(t)=-16t^{2}+32t + 5\), where \(a=-16\) and \(b = 32\)
The axis of symmetry is \(t=-\frac{32}{2\times(-16)}=\frac{-32}{-32}=1\). Since \(2>1\) and \(a=-16<0\) (parabola opens downwards), Bozo is going down at \(t = 2\) seconds.

• Part b

Substitute \(t = 2\) into \(f(t)=-16t^{2}+32t + 5\)

$$f(2)=-16\times2^{2}+32\times2+5=-16\times4+64 + 5=-64+64+5=5$$

Answer:

1.

• a. \(3\) seconds
• b. \(144\) feet

2.

• a. \(67\) feet
• b. Down

3.

• a. Down
• b. \(78\) feet

4.

• a. Down
• b. \(5\) feet