Verified Answer :
Answer:
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
- 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:
\(ax^{2} + bx + c, a
eq0\).
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.
A similar question is given at https://brainly.com/question/16858635