Question

Using Equations of Parabolas Quick Check
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Question
Jackie, who is 5 feet tall, throws a football into the air as if she were throwing it to someone. The path of the football can be expressed by the equation y=−0.05x2+0.75x+5
. Solve as if you were to graph the parabola to figure out how far from Jackie the ball landed.(1 point)
Responses

5 feet
5 feet

18.5 feet
18.5 feet

7.5 feet
7.5 feet

20 feet

Answer 1

To find out how far from Jackie the ball landed, we need to determine the x-coordinate where the football hits the ground, which occurs when \( y = 0 \). We start with the equation of the parabola:

\[
y = -0.05x^2 + 0.75x + 5
\]

Setting \( y \) to 0 to find the x-intercepts:

\[
0 = -0.05x^2 + 0.75x + 5
\]

To make it easier to solve, we can multiply the entire equation by -20 to eliminate the decimal:

\[
0 = x^2 - 15x - 100
\]

Now, this is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -15 \), and \( c = -100 \). We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Calculating the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-15)^2 - 4(1)(-100) = 225 + 400 = 625
\]

Now, substituting back into the quadratic formula:

\[
x = \frac{15 \pm \sqrt{625}}{2(1)} = \frac{15 \pm 25}{2}
\]

Calculating the two possible x-values:

1. \( x = \frac{15 + 25}{2} = \frac{40}{2} = 20 \)
2. \( x = \frac{15 - 25}{2} = \frac{-10}{2} = -5 \) (not applicable, as distance cannot be negative)

The only valid solution is \( x = 20 \).

Thus, the football landed **20 feet** away from Jackie.

The correct response is:

**20 feet**.