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Solution Set: 1≤ t ≤ 4.75
h(t)= 0
Factor out -4t
.LHS /(-4).=.RHS /(-4).
Use the Zero Product Property
(II): LHS+23=RHS+23
(II): .LHS /4.=.RHS /4.
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.
the rocket is at least 76 feet highcan 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.
Factor out -4
.LHS /(-4).=.RHS /(-4).
LHS+19=RHS+19
Use the Quadratic Formula: a = 4, b= - 23, c= 19
- (- a)=a
Calculate power and product
Subtract term
Calculate root
State solutions
Add and subtract terms
Calculate quotient
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.