Math

QuestionAn athlete releases a shot at 3535^{\circ}, modeled by f(x)=0.01x2+0.7x+5.4f(x)=-0.01 x^{2}+0.7 x+5.4. Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The path of the shot put is modeled by the quadratic function f(x)=0.01x+0.7x+5.4f(x)=-0.01 x^{}+0.7 x+5.4 . xx represents the horizontal distance in feet from the point of release3. f(x)f(x) represents the height of the shot put in feet

STEP 2

The maximum height of the shot put is given by the vertex of the parabola. 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 the coefficients a=0.01a = -0.01 and b=0.7b =0.7 into the formula to find the x-coordinate of the vertex.
x=b2a=0.72(0.01)x = -\frac{b}{2a} = -\frac{0.7}{2(-0.01)}

STEP 4

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

STEP 5

Substitute x=35x =35 into the function f(x)f(x) to find the maximum height of the shot put.
f(35)=0.01(35)2+0.7(35)+5.4f(35) = -0.01(35)^2 +0.7(35) +5.4

STEP 6

Calculate the maximum height of the shot put.
f(35)=0.01(35)2+0.(35)+5.4=18.15f(35) = -0.01(35)^2 +0.(35) +5.4 =18.15The maximum height of the shot put is18.15 feet and this occurs35 feet from the point of release.

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