Math

QuestionAn athlete releases a shot at 4040^{\circ} with height modeled by f(x)=0.01x2+0.8x+6.2f(x)=-0.01 x^{2}+0.8 x+6.2. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot can be modeled by the function f(x)=0.01x+0.8x+6.f(x)=-0.01 x^{}+0.8 x+6., where xx is the shot's horizontal distance in feet from its point of release. . We are asked to find the maximum height of the shot and the distance from the point of release where this maximum height occurs.

STEP 2

The maximum height of the shot occurs at the vertex of the parabola represented by the function f(x)f(x). The xx-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

Substitute the values of aa and bb from the given function into the formula to find the xx-coordinate of the vertex.
x=b2a=0.82(0.01)x = -\frac{b}{2a} = -\frac{0.8}{2(-0.01)}

STEP 4

Calculate the xx-coordinate of the vertex.
x=0.82(0.01)=40x = -\frac{0.8}{2(-0.01)} =40

STEP 5

Now that we have the xx-coordinate of the vertex, we can find the maximum height by substituting this value into the function f(x)f(x).
f(40)=0.01(40)2+0.8(40)+.2f(40) = -0.01(40)^2 +0.8(40) +.2

STEP 6

Calculate the maximum height.
f(40)=0.01(40)2+0.8(40)+6.2=22f(40) = -0.01(40)^2 +0.8(40) +6.2 =22The maximum height of the shot is22 feet, which occurs40 feet from the point of release.

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