Question

A certain water feature at a splash pad sprays water from a hole in the ground in the form of a parabola that can be expressed by the equation y=−34(x−2)2+3 . Solve as if you were to graph the equation. At which point would the water hit the ground?

Answer 1

To determine where the water hits the ground, we need to find the point at which \(y = 0\) (the ground level) in the given equation:

\[
y = -34(x - 2)^2 + 3
\]

Setting \(y = 0\):

\[
0 = -34(x - 2)^2 + 3
\]

Now we will solve for \(x\):

1. Subtract 3 from both sides:

\[
-3 = -34(x - 2)^2
\]

2. Divide by -34:

\[
\frac{3}{34} = (x - 2)^2
\]

3. Take the square root of both sides, remembering to consider both the positive and negative roots:

\[
x - 2 = \pm \sqrt{\frac{3}{34}}
\]

4. Solve for \(x\):

\[
x = 2 \pm \sqrt{\frac{3}{34}}
\]

Next, we can calculate the approximate values:

\[
\sqrt{\frac{3}{34}} \approx \sqrt{0.08824} \approx 0.297
\]

Thus, we can find the two potential \(x\) values where the water hits the ground:

\[
x \approx 2 + 0.297 \quad \text{and} \quad x \approx 2 - 0.297
\]

Calculating these:

1. \(x \approx 2 + 0.297 \approx 2.297\)
2. \(x \approx 2 - 0.297 \approx 1.703\)

Therefore, the water will hit the ground at approximately:

\[
x \approx 1.703 \quad \text{and} \quad x \approx 2.297
\]

So, the points where the water hits the ground are approximately \( (1.703, 0) \) and \( (2.297, 0) \).