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.

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

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.

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

If an equation has the form ax^2 + bx + c

The completed squared form will be

a(x + \frac{b}{2a})^2-a(\frac{b}{2a})^2 + c

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

The maximum (or minimum) point is given by (-\frac{b}{2a}, -a(\frac{b}{2a})^2 + c)

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

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.