Question

June kicked a small rock off a ledge that was meters above a crevasse f(x)= -4x^2 + 9
When will the rock hit the bottom of the crevasse 2 meters below ground level

Answer 1

To determine when the rock will hit the bottom of the crevasse (2 meters below ground level), we need to set up the equation for the function \( f(x) \).

The function given is:

\[
f(x) = -4x^2 + 9
\]

To find when the rock reaches 2 meters below ground level, we set \( f(x) = -2 \):

\[
-4x^2 + 9 = -2
\]

Now, we'll solve for \( x \):

1. Add 2 to both sides:

\[
-4x^2 + 9 + 2 = 0
\]

\[
-4x^2 + 11 = 0
\]

2. Rearranging gives:

\[
4x^2 = 11
\]

3. Divide both sides by 4:

\[
x^2 = \frac{11}{4}
\]

4. Take the square root of both sides:

\[
x = \pm \sqrt{\frac{11}{4}} = \pm \frac{\sqrt{11}}{2}
\]

Since we are concerned only with the positive value (time cannot be negative), we have:

\[
x = \frac{\sqrt{11}}{2}
\]

Thus, the rock will hit the bottom of the crevasse (2 meters below ground level) at:

\[
x \approx \frac{3.32}{2} \approx 1.66 \text{ meters}
\]

This gives the time in relation to the variable used here, assuming the variable \( x \) represents time. Therefore, the rock will hit the bottom of the crevasse approximately 1.66 meters (or whatever unit corresponds) after it was kicked off. Note that the original problem might have wanted an explicit time measurement which depends on additional context.