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QuestionAn athlete releases a shot modeled by f(x)=−0.02x2+1.7x+5.7f(x)=-0.02 x^{2}+1.7 x+5.7. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot is modeled by the quadratic function f(x)=−0.02x+1.7x+5.7f(x)=-0.02 x^{}+1.7 x+5.7 . The maximum height of the shot is the vertex of the parabola represented by the quadratic function3. The horizontal distance at which the maximum height occurs is the x-coordinate of the vertex

STEP 2

The vertex of a parabola represented by the quadratic function f(x)=ax2+bx+cf(x)=ax^{2}+bx+c is given by the formula for the x-coordinate of the vertex, hh, which is h=−b2ah=-\frac{b}{2a}.

STEP 3

We can find the x-coordinate of the vertex by substituting the coefficients aa and bb from our function into this formula.
h=−b2a=−1.72×−0.02h=-\frac{b}{2a}=-\frac{1.7}{2 \times -0.02}

STEP 4

Calculate the x-coordinate of the vertex.
h=−1.72×−0.02=42.h=-\frac{1.7}{2 \times -0.02}=42.

STEP 5

Now that we have the x-coordinate of the vertex, we can find the y-coordinate, which represents the maximum height of the shot. We can do this by substituting the x-coordinate of the vertex into our function.
f(h)=f(42.5)=−0.02×42.52+1.7×42.5+5.7f(h)=f(42.5)=-0.02 \times42.5^{2}+1.7 \times42.5+5.7

STEP 6

Calculate the y-coordinate of the vertex.
f(42.5)=−0.02×42.52+1.×42.5+5.=45.125f(42.5)=-0.02 \times42.5^{2}+1. \times42.5+5.=45.125The maximum height of the shot is45.125 feet, which occurs42.5 feet from the point of release.

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