Flights of leaping animals typically have parabolic paths. The figure illustrates a frog jump superimposed on a coordinate plane. The length of the leap is 9 feet, and the maximum height off the ground is 3a feet. Find a standard equation for the path of the frog. Assume a = 1.

Since the leap travels 9 feet, the function has zeros at x=0,9, so

y(x) = ax(x-9)

When x = 9/2,
y(9/2) = a(9/2)(-9/2) = -81/4 a

Why should we assume a=1?

If -81/4 a = 3a we are in trouble.

I think there's something wrong with the way the problem is stated. Either that, or (missing a diagram) my assumption that the height is zero at x=0,9 is wrong.

To find a standard equation for the path of the frog's leap, we can start by analyzing the given information. We know that the length of the leap is 9 feet, and the maximum height off the ground is 3a feet. Given that a = 1, the maximum height would be 3 feet.

First, let's consider the parabolic path of a frog's leap. A parabola can be represented by the equation y = ax^2 + bx + c, where a, b, and c are constants.

To find the equation, we need to determine the values of a, b, and c. We can do this by plugging in the coordinates of a point on the parabolic path into the equation and solving for the constants.

Considering the frog's leap, the coordinates of the highest point (the maximum height) can be represented as (x, 3), where x is the distance from the starting point.

Since the maximum height occurs at the midpoint of the leap (where the frog starts and lands), the x-coordinate of the highest point would be half of the leap's length. In this case, x = 9/2 = 4.5 feet.

Substituting these values into the equation, we have:
3 = a(4.5)^2 + b(4.5) + c

Since a = 1, this simplifies to:
3 = (4.5)^2 + b(4.5) + c
3 = 20.25 + 4.5b + c

Now, let's consider the starting point of the frog's leap. The frog starts at the origin, which means the coordinates of the starting point are (0, 0). Substituting these values into the equation, we have:
0 = a(0)^2 + b(0) + c

Since the starting point is at the origin, this simplifies to:
0 = c

Now, we have two equations:
3 = 20.25 + 4.5b
0 = c

We can solve the first equation for b:
3 - 20.25 = 4.5b
-17.25 = 4.5b
b = -3.83

Using the value of c = 0, we now have the equation:
y = ax^2 + bx + 0

Substituting the value of a = 1 and b = -3.83, the equation simplifies to:
y = x^2 - 3.83x

Therefore, the standard equation for the path of the frog's leap is y = x^2 - 3.83x.