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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 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.
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
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.
0 ≤ t ≤ 5.75