Mathematics, 18.08.2019 17:20 reginaldboyd28

The function y = -0.03(x - 14)^2 + 6 models the mump of a red kangaroo where x is the horizontal distance in meters and y is the vertical distance in meters for the height of the jump. what is the kangaroo's maximum height? how long is the kangaroo's jump?

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From the vertex of the quadratic equation, we find that: The kangaroos maximum height is of 6 meters.From the roots of the equation, we find that: The kangaroo's jump is 28.14 meters long.

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Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^{2} + bx + c$

It's vertex is the point $(x_{v}, y_{v})$

In which

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

Where

$\Delta = b^2-4ac$

If a<0, the vertex is a maximum point, that is, the maximum value happens at $x_{v}$ , and it's value is $y_{v}$ .

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Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

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The quadratic equation is:

y = -0.03(x - 14)^2 + 6

Placing in standard form:

y = -0.03(x^2 - 28x + 196) + 6

y = -0.03x^2 + 0.84x + 0.12

Thus, it has coefficients a = -0.03, b = 0.84, c = 0.12

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The kangaroo's maximum height is the y-value of the vertex, thus:

$\Delta = b^2 - 4ac = (0.84)^2 - 4(-0.03)(0.12) = 0.72$

$y_{v} = -\frac{\Delta}{4a} = -\frac{0.72}{4(-0.03)} = 6$

The kangaroos maximum height is of 6 meters.

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The length of the kangaroo's jump is the positive root. The roots are found at the values of x for which y = 0, thus, the solutions of the quadratic equation.

$x_{1} = \frac{-0.84 + \sqrt{0.72}}{2(-0.03)} = -0.14$

$x_{2} = \frac{-0.84 - \sqrt{0.72}}{2(-0.03)} = 28.14$

The kangaroo's jump is 28.14 meters long.

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Answer from: connienash95

Kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long

Step-by-step explanation:

Given : y = -0.03(x - 14)^2 + 6

To Find : What is the kangaroo's maximum height? How long is the kangaroo's jump?

Solution:

y = -0.03(x - 14)^2 + 6

x is the horizontal distance in meters

y is the vertical distance in meters for the height of the jump.

Substitute y = 0

0 = -0.03(x - 14)^2 + 6

0.03(x - 14)^2 = 6

$(x - 14)^2 = \frac{6}{0.03}$

(x - 14)^2 =200

$(x - 14) =\sqrt{200}$

(x - 14) =14.142

x =14.142+14

x =28.142

x≈ 28 m

Now the maximum height will be attained at mid point i.e. $\frac{28}{2} =14$

Now substitute x= 14

y = -0.03(14 - 14)^2 + 6

y = 6

So, kangaroo's maximum height is 6 m and the kangaroo's jump is 28 m long

Answer from: Quest

Ithink number to is 1 inch=5 miles
This is the picture of the homework

Answer from: Quest

Bc equals mb that should be your answer

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