If the path of a frog's jump is approximated by the equation

y = -0.25x2 + 5x, what is the height of the frog's jump?

What they are really asking is: Where is the vertex of this parabola?

If you know calculus, it is really easy. You look for where dy/dx = 0
dy/dx = -.5 x + 5
that is 0 when x = 10
then y = -100/4 + 50 or 25
If you do not use calculus then you must put this parabola in standard form by completing the square
-4y = x^2 -20 x
-4y + 100 = x^2 -20 x + 100
-4 (y-25) = (x-10)^2
vertex at (10,25)
as we knew

There are two ways to do this: calculus and completing the square. I don't know if they let you use calculus, so let's do it the other way. "y" is supposed to be the height. First find the value of x for which y is a maximum.

Rewrite as

y = -(x/2)^2 + 5x - (5)^2 + (5)^2 = 0
y = -[(x/2) - 5)]^2 + (5)^2 = 0
The maximum value of y occurs when x = 10, at which time the term in brackets os zero. The value of y them is 25, and that is the maximum.

Glad I agree with Damon!

To determine the height of the frog's jump, we need to find the maximum value of y in the given equation.

The equation y = -0.25x^2 + 5x represents a parabola because of the presence of x^2 term. In standard form, a parabola is represented as y = ax^2 + bx + c, where a, b, and c are constants.

In this case, a = -0.25, b = 5, and c = 0 (since there is no constant term).

To find the maximum height of the jump, we observe that the vertex of the parabola represents the highest point. The x-coordinate of the vertex can be found using the formula x = -b / (2a).

Substituting the values, we have x = -(5) / (2 * (-0.25)) = -5 / (-0.5) = 10.

Now, substitute this value of x into the equation y = -0.25x^2 + 5x to find the corresponding height (y-coordinate):

y = -0.25(10)^2 + 5(10) = -0.25(100) + 50 = -25 + 50 = 25.

Therefore, the height of the frog's jump is 25 units.