Question

A frog leaps 2 feet. The highest point in the jump is 6 inches. Assume the frog starts at (0, 0). Write a quadratic function in vertex form for the path of the jump. Specify the units you use.

Answer 1

Answer:

$\displaystyle f(x)=-\frac{1}{24}\left(x-12\right)^2+6$

Where f(x) models the height, in inches, of the frog's leap after it had leapt x inches.

Step-by-step explanation:

We are given that a frog leaps two feet, with the highest point in the jump being six inches.

Assuming the frog starts at (0,0), we want to find a quadratic function in vertex form for the path of the jump.

First, we will convert the length of the jump to inches. Two feet is equivalent to 24 inches.

Next, since the frog started at (0, 0), in order to jump a length of 24 inches, it must have ended at (0, 24). These two points will also be our two roots: x = 0 and x = 24. We can use the factored form of a quadratic:

$f(x)=a(x-p)(x-q)$

Where p and q are the roots.

Substitute in the roots:

$f(x)=a(x-0)(x-24)=ax(x-24)$

Now, since the frog reached a maximum height of six inches, the y-coordinate of the function at its vertex point is 6.

Since a parabola is symmetrical along its axis of symmetry, the axis of symmetry is always halfway between the two roots.

Therefore, the x-coordinate of our vertex is:

$\displaystyle x=\frac{24-0}{2}=12$

So, when x = 12, f(x) = 6. Substitute:

$6=a(12)(12-24)$

Solve for a:

$\displaystyle- 144a=6\Rightarrow a=-\frac{1}{24}$

Therefore, our function is:

$\displaystyle f(x)=-\frac{1}{24}x(x-24)$

Now, we can place this into vertex form:

$\displaystyle f(x)=a(x-h)^2+k$

Where a is the leading coefficient and (h, k) is the vertex.

Our leading coefficient is -1/24, and the vertex is (12, 6). Hence, our function is:

$\displaystyle f(x)=-\frac{1}{24}\left(x-12\right)^2+6$

Where f(x) models the height, in inches, of the frog's leap after it had leapt x inches.