Question

Jada dives off a cliff into the ocean. The vertical path of her dive, in feet, is modeled by the function f(x)=−0.1(x−3)2+10
𝑓
(
𝑥
)
=
−
0.1
(
𝑥
−
3
)
2
+
10
, where x is the horizontal distance and f(x)
𝑓
(
𝑥
)
is the vertical distance. To capture her experience, Jada asks a friend to record her on a video camera. The camera’s view is modeled by the function g(x)=0.67x+3
𝑔
(
𝑥
)
=
0.67
𝑥
+
3
, where x is the horizontal distance, in feet, and g(x)
𝑔
(
𝑥
)
is the vertical distance, in feet, that the camera can capture. Graph the equations to determine whether the camera will be able to capture Jada’s jump before she hits the water. If so, how far above the surface of the water, to the nearest integer, will the camera capture her jump?(1 point)
Responses

3 feet
3 feet

8 feet
8 feet

7 feet
7 feet

The camera will not capture her jump

Answer 1

To determine whether the camera can capture Jada's jump before she hits the water, we need to analyze the two functions \( f(x) \) and \( g(x) \).

1. **Model of Jada's Dive**:
\[
f(x) = -0.1(x - 3)^2 + 10
\]
This is a quadratic function that opens downwards. The vertex form of the parabola indicates that the peak height (maximum vertical distance) occurs at \( x = 3 \) with a maximum height of 10 feet.

2. **Model of the Camera’s View**:
\[
g(x) = 0.67x + 3
\]
This is a linear function with a slope of 0.67 and a y-intercept of 3 feet.

3. **Finding Intersection Points**:
To find if the camera captures Jada's dive, we set \( f(x) = g(x) \):
\[
-0.1(x - 3)^2 + 10 = 0.67x + 3
\]

Rearranging gives:
\[
-0.1(x - 3)^2 - 0.67x + 7 = 0
\]
We can multiply through by -10 to eliminate the fraction:
\[
(x - 3)^2 + 6.7x - 70 = 0
\]
Expanding \( (x - 3)^2 \):
\[
x^2 - 6x + 9 + 6.7x - 70 = 0
\]
Simplifying it:
\[
x^2 + 0.7x - 61 = 0
\]

4. **Using the Quadratic Formula**:
The roots can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = 0.7, c = -61 \):
\[
x = \frac{-0.7 \pm \sqrt{(0.7)^2 - 4 \cdot 1 \cdot (-61)}}{2 \cdot 1}
\]
\[
= \frac{-0.7 \pm \sqrt{0.49 + 244}}{2}
\]
\[
= \frac{-0.7 \pm \sqrt{244.49}}{2}
\]
\[
= \frac{-0.7 \pm 15.64}{2}
\]
Two possible solutions for \( x \):
\[
x_1 = \frac{14.94}{2} \approx 7.47
\]
\[
x_2 = \frac{-16.34}{2} \approx -8.17 \quad (\text{not relevant since distances can't be negative})
\]

5. **Finding the Height at Intersection**:
We use \( x = 7.47 \) to find the corresponding \( f(x) \):
\[
f(7.47) = -0.1(7.47 - 3)^2 + 10
\]
\[
= -0.1(4.47)^2 + 10
\]
\[
= -0.1(19.97) + 10
\]
\[
= -1.997 + 10 \approx 8.003 \text{ feet}
\]

Thus, the camera will capture Jada's jump at approximately **8 feet** above the surface of the water.

**Final Answer**:
8 feet