[rachh]

Tanay kicks a soccer ball during a game. The height of the ball, in feet, can be modeled by the function f(x)= -16xto the second power + 48x, where x is the time in seconds after she kicks the ball. Graph the function. Find the maximun height of the ball and how long it takes the ball to reach that height.

[Capuchin]

Have you graphed it? We can't exactly graph it for you...

[Zazonker]

Assuming you are new to graphing and having trouble with that. You'd start with x on the horizontal axis and f(x) on the verticle. You calculate the value of f(x) for a few values of x just by plugging them into the equation. If you try x = 1, x=2 and x=3 you get 32, 32 and 0 respectively.

Visualize the soccer ball in flight - it certainly didn't jump instantly to 32 feet, stay there for a couple of seconds and then drop instantly to the ground. From that you know that you need smaller increments than whole seconds, maybe say tenths of a second. And the high point was before 3 seconds - quite likely between 1 and 2 since the only way the ball could be at 32 at both times was if it was going up at 1 second and coming down at 2 seconds.

From this, I think you should be able to solve it.

[Unknown008]

Hint : complete the square to find your reference points to draw your graph.

If an equation has the form

The completed squared form will be



Setting x to zero and finding the resulting value will give you the y intercept.

The maximum (or minimum) point is given by

[Zazonker]

Seems you are making this a bit more difficult than it is. The Y intercept can be "calculated" just be setting x to zero in the original equation. No need to complete the square.

F(x) = -16 x ^2 + 48 x
If x = 0, f(x) = 0

The max can be determined by setting the first derivative to zero.
That is:
-32x + 48 = 0
x= 48/32 = 1.5

[Unknown008]

Yeah, I know, but I don't know whether the OP has been learning derivatives...

The formula really seems long, but when you plug in the numbers, it's lots shorter and simpler.