Math

QuestionAn athlete releases a shot at 5555^{\circ} with height modeled by f(x)=0.02x2+1.4x+6.1f(x)=-0.02 x^{2}+1.4 x+6.1. 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)=0.02x+1.4x+6.1f(x)=-0.02 x^{}+1.4 x+6.1 . xx represents the horizontal distance, in feet, from the point of release3. We are looking for the maximum height of the shot put and the distance from the point of release at which this occurs

STEP 2

The maximum height of the shot put occurs at the vertex of the parabola represented by the equation f(x)=0.02x2+1.4x+6.1f(x)=-0.02 x^{2}+1.4 x+6.1. 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

Substitute the values of aa and bb from the equation f(x)=0.02x2+1.x+6.1f(x)=-0.02 x^{2}+1. x+6.1 into the formula b2a-\frac{b}{2a} to find the x-coordinate of the vertex.
x=b2a=1.2×0.02x = -\frac{b}{2a} = -\frac{1.}{2 \times -0.02}

STEP 4

Calculate the x-coordinate of the vertex.
x=1.42×0.02=35x = -\frac{1.4}{2 \times -0.02} =35

STEP 5

Substitute x=35x=35 into the equation f(x)=0.02x2+1.4x+.1f(x)=-0.02 x^{2}+1.4 x+.1 to find the maximum height of the shot put.
f(35)=0.02×352+1.4×35+.1f(35) = -0.02 \times35^{2} +1.4 \times35 +.1

STEP 6

Calculate the maximum height of the shot put.
f(35)=0.02×352+1.4×35+6.1=31.1f(35) = -0.02 \times35^{2} +1.4 \times35 +6.1 =31.1The maximum height of the shot put is31.1 feet, which occurs35 feet from the point of release.

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