Math

QuestionAn athlete releases a shot at 6060^{\circ}, modeled by f(x)=0.02x2+1.7x+6.1f(x)=-0.02 x^{2}+1.7 x+6.1. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The path of the shot put can be modeled by the quadratic function f(x)=0.02x+1.7x+6.1f(x)=-0.02x^{}+1.7x+6.1 . The maximum height of the shot put corresponds to the maximum value of the function f(x)f(x)3. The horizontal distance from the point of release at which the maximum height occurs corresponds to the xx-value at which the maximum value of f(x)f(x) occurs

STEP 2

The maximum or minimum of a quadratic function ax2+bx+cax^{2}+bx+c occurs at the vertex of the parabola. The xx-coordinate of the vertex can be found using the formula 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=1.72(0.02)x = -\frac{b}{2a} = -\frac{1.7}{2(-0.02)}

STEP 4

Calculate the xx-coordinate of the vertex.
x=1.72(0.02)=42.x = -\frac{1.7}{2(-0.02)} =42.This means that the maximum height occurs42. feet from the point of release.

STEP 5

Substitute the xx-coordinate of the vertex back into the function to find the maximum height.
f(x)=0.02(42.5)2+1.7(42.5)+.1f(x) = -0.02(42.5)^{2}+1.7(42.5)+.1

STEP 6

Calculate the maximum height.
f(x)=0.02(42.5)2+1.(42.5)+6.1=38.375f(x) = -0.02(42.5)^{2}+1.(42.5)+6.1 =38.375The maximum height of the shot is38.375 feet.

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