Question

The function f(x) = −0.002x2 + 0.66x models the path of a softball that is hit for a home run, where f(x) gives the height of the ball and x gives the distance from where it is hit in feet.

a. How far does the ball travel before hitting the ground?
b. How high does the ball go?
c. What is a reasonable domain and range for a home run modeling function?

Answer 1

Answer:

a) $0 = x(-0.002 x +0.66)$

So then we see that x =0 or $x = \frac{0.66}{0.002}=330$

So then we can say that the ball travels 330 ft before hitting the ground

b) For this case we want to find the maximum of this function and since the coeffcient for the x^2 term is negative if we find the vertex we find the maximum, the x coordinate for the vertex from a quadratic function $y= ax^2 +bx+c$ is given by:

$V_x = -\frac{b}{2a}= -\frac{0.66}{2*(-0.002)}=165 ft$

And for the y coordinate we have:

$f(V_x) = -0.002(165^2) +0.66(165) = 54.45 ft$

So then the maximum height for this case would be 54.45.

c) For this case the function makes sense for values of $f(x) \geq 0$

And for this function we satisfy this condition with domain $x \in [0,330]$ and the range $y \in [0,54.45]$

Step-by-step explanation:

For this case w ehave the following function:

$f(x) = -0.002 x^2 +0.66 x$

Where f(x) gives the height of the ball and x gives the distance from where it is hit in feet

Part a

For this case we want to find when f(x) =0 so we can do this:

$0 = -0.002 x^2 +0.66 x$

We can take common factor x like this:

$0 = x(-0.002 x +0.66)$

So then we see that x =0 or $x = \frac{0.66}{0.002}=330$

So then we can say that the ball travels 330 ft before hitting the ground

Part b

For this case we want to find the maximum of this function and since the coeffcient for the x^2 term is negative if we find the vertex we find the maximum, the x coordinate for the vertex from a quadratic function $y= ax^2 +bx+c$ is given by:

$V_x = -\frac{b}{2a}= -\frac{0.66}{2*(-0.002)}=165 ft$

And for the y coordinate we have:

$f(V_x) = -0.002(165^2) +0.66(165) = 54.45 ft$

So then the maximum height for this case would be 54.45.

Part c

For this case the function makes sense for values of $f(x) \geq 0$

And for this function we satisfy this condition with domain $x \in [0,330]$ and the range $y \in [0,54.45]$