The function h(x)=-0.03(x-14)^2+6 models the jump of a red kangaroo, where x is the horizontal distance traveled (in feet) and h(x) is the height (in feet). When the kangaroo jumps from a higher location, it lands 5 feet farther away. Write a function that models the second jump. Round values in your equation to the nearest hundredth, if necessary.

Respuesta :

A curve that is representable with a quadratic function f(x) = a·x² + b·x + c, is a parabola

  • The function h(x) = -0.03·(x - 14)² + 10.83 models the second jump of the red kangaroo

The reason the above function is correct is as follows:

The known parameters are;

The given function that models the jump of the kangaroo is h(x) = -0.03·(x - 14)² + 6

The horizontal distance traveled (in feet) = x

The height (in feet) = h(x)

The position the kangaroo lands when it jumps from a higher location = 5 feet farther away

Required:

To write a function that models the second jump

Solution:

Let  y₀ represent the added height, we have;

The coordinates of the vertex of the parabola given in vertex form, y = a·(x - h)² + k is (h, k)

Therefore;

The coordinates of vertex of h(x) = -0.03·(x - 14)² + 6, is (14, 6)

The highest point of the parabola is 6 feet, and 14 feet is the midpoint of the parabola

Therefore, by symmetry of a parabola, we have;

The maximum range of the parabola is 2 × 14 feet = 28 feet

From the new height, y₀, the new range is 28 feet + 5 feet = 33 feet

From the new height, the second jump, we have;

The height of the parabola, h(x) = y₀ + (-0.03·(x - 14)² + 6)

The height when the kangaroo lands at 33 feet away is h(33) = 0 feet

The initial height of the kangaroo = y₀ feet

Therefore, we get;

h(33) = 0 = y₀ + (-0.03·(33 - 14)² + 6) = y - 4.83

y₀ - 4.83 = 0

y₀ = 4.83

The initial height of the kangaroo, y₀ = 4.83

Therefore, for the second jump, we have;

h(x) = 4.83 + (-0.03·(x - 14)² + 6) = -0.03·(x - 14)² + 10.83

Which gives;

The function that models the second jump is h(x) = -0.03·(x - 14)² + 10.83

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