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 , where x is the horizontal distance and 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 , where x is the horizontal distance, in feet, and 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 7 feet 7 feet The camera will not capture her jump. The camera will not capture her jump. 8 feet 8 feet 3 feet 3 feet Skip to navigation

1 answer

To determine whether the camera can capture Jada's jump, we need to find the intersection of the two equations \( f(x) = -0.1(x - 3)^2 + 10 \) for her dive and \( g(x) = 0.67x + 3 \) for the camera's view.

  1. Find the equations:

    • Jada's dive: \[ f(x) = -0.1(x - 3)^2 + 10 \]
    • Camera view: \[ g(x) = 0.67x + 3 \]
  2. Set the equations equal to each other to find the intersection points: \[ -0.1(x - 3)^2 + 10 = 0.67x + 3 \]

  3. Rearranging the equation: \[ -0.1(x - 3)^2 - 0.67x + 7 = 0 \]

  4. Clear the equation: Let's first expand \( -0.1(x - 3)^2 \): \[ -0.1(x^2 - 6x + 9) = -0.1x^2 + 0.6x - 0.9 \] Now substitute it into the equation: \[ -0.1x^2 + 0.6x - 0.9 - 0.67x + 7 = 0 \]

  5. Combine like terms: \[ -0.1x^2 - 0.07x + 6.1 = 0 \]

  6. Multiply by -10 to eliminate decimals: \[ x^2 + 0.7x - 61 = 0 \]

  7. Now we can use the quadratic formula to find x: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 0.7, c = -61 \).

    \[ b^2 - 4ac = 0.7^2 - 4(1)(-61) = 0.49 + 244 = 244.49 \] \[ x = \frac{-0.7 \pm \sqrt{244.49}}{2} \] \[ x = \frac{-0.7 \pm 15.634}{2} \]

    This gives us two potential solutions for \( x \): \[ x_1 = \frac{14.934}{2} \approx 7.467 \] \[ x_2 = \frac{-16.334}{2} \approx -8.167 \quad(\text{not relevant since it is negative}) \]

  8. Find \( f(x) \) and \( g(x) \) at \( x \approx 7.467 \):

    • To find the height of Jada at this x value: \[ f(7.467) = -0.1(7.467 - 3)^2 + 10 \] \[ f(7.467) = -0.1(4.467)^2 + 10 \approx -0.1(19.94) + 10 \approx -1.994 + 10 = 8.006 \]

    • To find the camera’s altitude: \[ g(7.467) = 0.67(7.467) + 3 \approx 5.007 + 3 = 8.007 \]

Both values are approximately 8 feet. Therefore, the camera captures Jada just as she is about to hit the water.

Conclusion: The camera will capture Jada's jump, and she will be approximately 8 feet above the water when captured. The answer to the query is 8 feet.