Question

To begin a football game, the kicker has to kick off. The football follows the path h(x) = −130x(x−60)%0D%0Aℎ%0D%0A(%0D%0A%0D%0A)%0D%0A %0D%0A=%0D%0A %0D%0A−%0D%0A1%0D%0A30%0D%0A%0D%0A(%0D%0A%0D%0A−%0D%0A60%0D%0A)%0D%0A , where h(x)%0D%0Aℎ%0D%0A(%0D%0A%0D%0A)%0D%0A is the height of the football and x is the horizontal distance it has traveled in yards.%0D%0A%0D%0AAnswer the questions below:%0D%0A%0D%0A(1 point)%0D%0A1) What is the y-intercept? Write your answer as an ordered pair. %0D%0A%0D%0A2) At what time is the football at the highest height? seconds.%0D%0A%0D%0A3) The ball will have traveled yards before hitting the ground.

Answer 1

To analyze the function \( h(x) = -130x(x - 60) \), we can address the questions step by step.

### Question 1: Y-intercept
The y-intercept occurs when \( x = 0 \). So we calculate \( h(0) \):

\[
h(0) = -130(0)(0 - 60) = -130(0)(-60) = 0
\]

The y-intercept is \( (0, 0) \).

### Question 2: Time at Highest Height
The function \( h(x) \) represents a quadratic equation in the form \( h(x) = ax^2 + bx + c \). The highest point (the vertex of the parabola) occurs at \( x = -\frac{b}{2a} \).

Here, we can rewrite the function as:

\[
h(x) = -130(x^2 - 60x) = -130x^2 + 7800x
\]

From this, we identify \( a = -130 \) and \( b = 7800 \).

Calculating the vertex:

\[
x = -\frac{b}{2a} = -\frac{7800}{2(-130)} = \frac{7800}{260} = 30
\]

So, the football reaches its highest height at \( x = 30 \) yards.

### Question 3: Distance Traveled Before Hitting the Ground
The football hits the ground when \( h(x) = 0 \). This occurs at the two points found by setting \( -130x(x - 60) = 0 \):

1. \( -130x = 0 \) → \( x = 0 \)
2. \( x - 60 = 0 \) → \( x = 60 \)

Thus, the ball will have traveled **60 yards** before hitting the ground.

### Summary of Answers:
1) The y-intercept is \( (0, 0) \).
2) The football is at its highest height at \( x = 30 \) yards.
3) The ball will have traveled **60 yards** before hitting the ground.