Math

QuestionAn athlete releases a shot at 3535^{\circ} with height modeled by f(x)=0.01x2+0.7x+5.8f(x)=-0.01 x^{2}+0.7 x+5.8. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot put is modeled by the function f(x)=0.01x+0.7x+5.8f(x)=-0.01 x^{}+0.7 x+5.8, where xx is the horizontal distance in feet from the point of release and f(x)f(x) is the height in feet. . The shot put is released at an angle of 3535^{\circ}.

STEP 2

The maximum height of the shot put can be found by finding the vertex of the parabola represented by the function f(x)=0.01x2+0.7x+5.8f(x)=-0.01 x^{2}+0.7 x+5.8. The x-coordinate of the vertex of a parabola given by the equation f(x)=ax2+bx+cf(x)=ax^{2}+bx+c is given by b2a-\frac{b}{2a}.

STEP 3

Plug in the values for aa and bb from the function f(x)=0.01x2+0.7x+5.8f(x)=-0.01 x^{2}+0.7 x+5.8 into the formula b2a-\frac{b}{2a} to find the x-coordinate of the vertex.
x=b2a=0.72×0.01x = -\frac{b}{2a} = -\frac{0.7}{2 \times -0.01}

STEP 4

Calculate the x-coordinate of the vertex.
x=0.72×0.01=35x = -\frac{0.7}{2 \times -0.01} =35

STEP 5

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs. Therefore, the maximum height occurs35 feet from the point of release.

STEP 6

To find the maximum height of the shot put, plug in the x-coordinate of the vertex into the function f(x)=0.01x2+0.x+5.8f(x)=-0.01 x^{2}+0. x+5.8.
f(x)=0.01x2+0.x+5.8=0.01(35)2+0.(35)+5.8f(x) = -0.01 x^{2}+0. x+5.8 = -0.01(35)^{2}+0.(35)+5.8

STEP 7

Calculate the maximum height of the shot put.
f(x)=0.01(35)2+0.7(35)+5.=17.75f(x) = -0.01(35)^{2}+0.7(35)+5. =17.75The maximum height of the shot put is17.75 feet and this occurs35 feet from the point of release.

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