Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 121 Page 526

Practice makes perfect
a When the rocket is on the ground, the function has a value of 0. Therefore, we can solve for the t-intercepts of the function by setting h(t) equal to 0 and using the Zero Product Property.
h(t)=-16t^2+92t
0=-16t^2+92t
â–Ľ
Solve for t
0=-4t(4t-23t)
0=t(4t-23)
lct=0 & (I) 4t-23=0 & (II)
lt=0 4t=23
lt_1=0 t_2=5.75
The function has two t-intercepts. One is at t=0, which is when the rocket is first launched, and the other is at t=5.75, when it lands. This means that the rocket reaches the ground 5.75 seconds after it is launched.
b We want to find how long the rocket is at a height of at least 76 feet. Since the height of the rocket is represented by h(t), the statement that the rocket is at least 76 feet high can be written as the inequality h(t)≥ 76. Let's replace the left-hand side of this inequality with the right-hand side of the function given in Part A.
-16t^2+92t≥ 76 To solve this inequality, we will start by treating it as an equation and solve it with the Quadratic Formula. In order to do that, though, we must first collect all terms on one side of the equation.
-16t^2+92t=76
-4(4t^2-23t)=-4(-19)
4t^2-23t=-19
4t^2-23t+19=0
Now we can solve the equation with by using the Quadratic Formula.
4t^2-23t+19=0
t=-( -23) ± sqrt(( - 23)^2-4( 4)( 19))/2 * 4
â–Ľ
Simplify right-hand side
t=23 ± sqrt((- 23)^2-4(4)(19))/2 * 4
t=23 ± sqrt(529-304)/8
t=23 ± sqrt(225)/8
t=23 ± 15/8
lct= (23-15)8 & (I) t= (23+15)8 & (II)
lct= 88 & (I) t= 388 & (II)
lt_1=1 t_2=4.75
We got two solutions, t=1 and t=4.75. Let's plot these on a number line. Since the inequality is non-strict, the values will be included in the solution set.

To figure out when the inequality is true, we will test a point that is less than 1, a second point that is between 1 and 4.75, and a third point that is greater than 4.75. If the value produces a true statment, we will shade in that part of the number line. Let's use 0, 2, and 6.

t - 16t^2+92t ≥ 76 Evaluate True?
0 - 16 0^2+92 0 ? ≥ 76 0≱ 76 *
2 - 16 2^2+92 2 ? ≥ 76 120≥ 76 ✓
6 - 16 6^2+92 6 ? ≥ 76 -24≱ 76 *

When t is between 1 and 4.75, the inequality is true. Let's shade the center section of the number line.

This means the rocket is at least 76 feet in the air from 1 to 4.75 seconds after it was launched.

c From Part B, we know that the function describing the rocket is at least 76 feet from 1 to 4.75 seconds after it was launched. Let's calculate how long that is. 4.75-1=3.75 The rocket was at least 76 feet high for 3.75 seconds. Therefore, Chad's rocket does not qualify.
d The domain tells us what values we can substitute for the independent variable in our function. Since the rocket cannot have negative values of t and it cannot go below the ground, we have the following domain.

0 ≤ t ≤ 5.75