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QuestionAn athlete releases a shot at 70∘70^{\circ} with height modeled by f(x)=−0.05x2+2.7x+6.2f(x)=-0.05 x^{2}+2.7 x+6.2. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put, f(x)f(x), is modeled by the equation f(x)=−0.05x+.7x+6.f(x)=-0.05 x^{}+.7 x+6.. . xx is the shot's horizontal distance, in feet, from its point of release.
3. We are looking for the maximum height of the shot and the distance from its point of release at which this occurs.

STEP 2

The maximum height of the shot occurs at the vertex of the parabola represented by the quadratic function f(x)=−0.05x2+2.7x+6.2f(x)=-0.05 x^{2}+2.7 x+6.2. The x-coordinate of the vertex of a parabola given by f(x)=ax2+bx+cf(x)=ax^2+bx+c is given by −b2a-\frac{b}{2a}.

STEP 3

Substitute a=−0.05a=-0.05 and b=2.7b=2.7 into the formula −b2a-\frac{b}{2a} to find the x-coordinate of the vertex.
x=−b2a=−2.72(−0.05)x = -\frac{b}{2a} = -\frac{2.7}{2(-0.05)}

STEP 4

Calculate the x-coordinate of the vertex.
x=−2.72(−0.05)=27x = -\frac{2.7}{2(-0.05)} =27

STEP 5

Substitute x=27x=27 into the function f(x)=−0.05x2+2.7x+.2f(x)=-0.05 x^{2}+2.7 x+.2 to find the maximum height of the shot.
f(27)=−0.05(27)2+2.7(27)+.2f(27) = -0.05(27)^2 +2.7(27) +.2

STEP 6

Calculate the maximum height of the shot.
f(27)=−0.05(27)2+2.(27)+6.2=36.85f(27) = -0.05(27)^2 +2.(27) +6.2 =36.85The maximum height of the shot is 36.8536.85 feet, which occurs 2727 feet from the point of release.

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