Having trouble with evaluating and solving quadratic functions

Question

Evaluating and Solving Quadratic Functions
A rocket is launched from a tower at an initial velocity of 108 feet per second. The height (in feet) of the rocket after tt seconds is given by the function h(t)=−16t2+108t+44h(t)=-16t2+108t+44. Use this function to answer the following questions. Round answers to the nearest hundredth as needed.
Determine the height of the launching tower.		The launching tower was  feet high.
After how many seconds does the rocket reach its highest point?		The rocket reaches its highest point after  seconds.
How high is the rocket at its highest point?		The rocket is  feet above the ground at its highest point.
After how many seconds does the rocket hit the ground?		The rocket will hit the ground  seconds after it is launched.
How long does it take for the rocket to reach a height of 33 feet above the ground?		It will take  seconds for the the rocket to reach a height of 33 feet above the ground.
Determine the practical domain of this function. Write your answer in Interval Notation		Practical Domain:
Determine the practical range of this function. Write your answer in Interval Notation		Practical Range:

Answer 1

If the height of the rocket is modeled by:

h(t) = -16t² + 108t + 44

Then, the height of the tower would be the same as the height of the rocket at t = 0 seconds (at the exact time it took off, the rocket is at the top of the tower)

For t = 0: h(0) = -16(0)² + 108(0) + 44 = 44 feet

The rocket follows a parabolic shape (a parabola is the shape of a quadratic). The highest point of a parabola is called its vertex. To find the "x-coordinate" of the parabola f(x) = ax² + bx + c, we use x = -b/(2a) so to find the "t-coordinate" of our parabola h(t) = -16t² + 108t + 44 we use t = -108/(2(-16)) = -108/-32 = 3.375. Meaning, the rocket reaches it's maximum height at t = 3.375 seconds. To find out what that height is, plug in t = 3.375 and see what you get:

For t = 3.375: h(3.375) = -16(3.375)² + 108(3.375) + 44 = 226.25 feet

To find out the time that the rocket hits the ground, you must realize that "the ground" is when the height is zero meaning h(t) = 0 so:

0 = -16t² + 108t + 44 but to make this easier, divided both sides by 4:

0 = -4t² + 27t + 11

Use the Quadratic Formula:

For ax2 + bx + c = 0, x = [-b ± √(b² - 4ac)]/(2a)

So t = [-27 ± √(27² - 4(-4)(11))]/(2(-4)) = [-27 ± √(729 + 176)]/(-8) = (-27 ± √905)/-8 = (-27 ± 30.083)/-8

Meaning that t = (-27 + 30.083)/-8 or t = (-27 - 30.083)/-8 which is t = -0.385 seconds or t = 7.135 seconds

We can through out the negative answer meaning the rocket hits the ground after 7.135 seconds.

To find the time when the rocket reaches 33 feet, we plug in a height of 33 into our equation:

33 = -16t² + 108t + 44 and solve for t by subtracting 33 from both sides and going through the same process to solve as we did above:

0 = -16t² + 108t + 11

t = [-108 ± √(108² - 4(-16)(11))]/(2(-16)) = [-108 ± √(11664 + 704)]/(-32) = [-108 ± √12368]/(-32) = [-108 ± 111.212]/(-32) so t = (-108 + 111.212)/-32 or t = (-108 - 111.212)/-32 meaning t = -0.1 seconds or t = 6.85 seconds. We can again throw out the negative answer so at t = 6.85 seconds the rocket height is 33 feet.

The practical domain (possible values of "t" are t > 0 and t < 7.135 ("take off" time to "hit the ground" time). To write this in interval notation we say domain: [0, 7.135]

The practical range (possible values of "h" are h > 0 and h < 226.25 "hit the ground" height to max height). To write this in interval notation we say range: [0, 226.25]

Answer 2

h(t) =-16t^2 +108t +44

ho= 44 feet = tower height

h'(t)=v(t) =-32t +108=0

t = 108/32 = 27/8 = 3 3/8 = 3.375 seconds to reach max height

max height = h(27/8) = -16(27/8)^2 + 108(27/8) + 44

= -182.24+ 354.5 +44= 182.25+44 = 226.25 feet maximum

it hits ground when h(t)=0=-16t^2 + 108t +44

16t^2-108t-44 =0

4t^2 -27t -11 = 0

t = 28/8 + (1/8)sqr(27^2+16(11))

= (28+sqr905)/8 = about 7.26 seconds to hit ground

it reaches 33 feet above ground when

h(t)=33 = -16t^2 +108t +44

16t^2 -108t -11 = 0

t= 108/32 +(1/32)sqr(108^2+64(11))

= (108 +sqr12368)/32

= about 6.85 seconds

domain [0, 7.26] in seconds

range [0, 226.25] in feet