Question

Velia hits a baseball toward the outfield fence. The function h=−16t2+80t+4 models the height of the ball, h, in feet, with respect to time, t, in seconds. What is the reasonable domain and range of the polynomial function?(1 point)
Responses

Domain: [0,∞);

Range: [0,104]

Domain: left bracket 0 comma infinity right parenthesis ; Range: left bracket 0 comma 104 right bracket

Domain: [0,5.05];

Range: [0,∞)

Domain: left bracket 0 comma 5 point 0 5 right bracket ; Range: left bracket 0 comma infinity right parenthesis

Domain: (−∞,∞);

Range: (−∞,∞)

Domain: open paren negative infinity comma infinity close paren ; Range: open paren negative infinity comma infinity close paren

Domain: [0,5.05];

Range: [0,104]

Answer 1

To determine the reasonable domain and range of the function \( h = -16t^2 + 80t + 4 \), let's analyze the problem.

The function describes the height of a baseball over time, which implies:

1. **Domain**: The time \( t \) cannot be negative, as you cannot have negative time in this context. Hence, the domain starts from \( t = 0 \) and goes to when the baseball hits the ground. We need to find when \( h(t) = 0 \) to determine the upper limit of \( t \).

To find when the baseball hits the ground, we can solve:
\[
-16t^2 + 80t + 4 = 0
\]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -16 \), \( b = 80 \), and \( c = 4 \):
\[
t = \frac{-80 \pm \sqrt{80^2 - 4 \cdot (-16) \cdot 4}}{2 \cdot (-16)}
\]
\[
= \frac{-80 \pm \sqrt{6400 + 256}}{-32}
\]
\[
= \frac{-80 \pm \sqrt{6656}}{-32}
\]
\[
= \frac{-80 \pm 81.64}{-32}
\]
This gives two potential values for \( t \):
\[
t \approx \frac{1.64}{-32} \quad \text{and} \quad t \approx \frac{-161.64}{-32}
\]
The positive value \( t \) will be around \( 5.05 \) seconds after calculation.

Therefore, the reasonable domain is:
\[
[0, 5.05]
\]

2. **Range**: To find the range, we need to determine the maximum height of the baseball. The height of the ball is a quadratic function that opens downwards (since the coefficient of \( t^2 \) is negative). We can find the vertex (maximum height) using the formula \( t = -\frac{b}{2a} \):
\[
t = -\frac{80}{2 \times -16} = \frac{80}{32} = 2.5
\]

Now substituting \( t = 2.5 \) back into the height function to get the maximum height:
\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 4
\]
\[
= -16(6.25) + 200 + 4
\]
\[
= -100 + 200 + 4 = 104 \quad \text{feet}
\]

The height \( h \) cannot be negative, so the reasonable range is:
\[
[0, 104]
\]

Combining both findings, the reasonable domain and range for the function are:
- **Domain**: [0, 5.05]
- **Range**: [0, 104]

Thus, the correct response from the options given is:
**Domain: [0, 5.05]; Range: [0, 104]**.