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QuestionA ball is thrown from 8 feet high. Its height is modeled by f(x)=−0.2x2+1.7x+8f(x)=-0.2 x^{2}+1.7 x+8. Find the maximum height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x), is given by the quadratic equation f(x)=−0.x+1.7x+8f(x)=-0. x^{}+1.7 x+8. . xx is the horizontal distance in feet from where the ball was thrown.
3. We need to find the maximum height of the ball and the distance from where it was thrown.

STEP 2

The maximum height of the ball is given by the vertex of the parabola represented by the equation f(x)=−0.2x2+1.7x+8f(x)=-0.2 x^{2}+1.7 x+8. 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}.
xvertex=−b2ax_{vertex} = -\frac{b}{2a}

STEP 3

Plug in the values for aa and bb from the equation f(x)=−0.2x2+1.7x+8f(x)=-0.2 x^{2}+1.7 x+8 to find the x-coordinate of the vertex.
xvertex=−1.72(−0.2)x_{vertex} = -\frac{1.7}{2(-0.2)}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=−1.7−0.4=4.25x_{vertex} = -\frac{1.7}{-0.4} =4.25

STEP 5

The maximum height of the ball is given by the y-coordinate of the vertex, which is f(xvertex)f(x_{vertex}). Substitute xvertexx_{vertex} into the equation f(x)f(x) to find the maximum height.
f(xvertex)=−0.2(4.25)2+1.7(4.25)+8f(x_{vertex}) = -0.2 (4.25)^{2}+1.7 (4.25)+8

STEP 6

Calculate the maximum height of the ball.
f(xvertex)=−0.2(4.25)2+1.(4.25)+8=11.425f(x_{vertex}) = -0.2 (4.25)^{2}+1. (4.25)+8 =11.425The maximum height of the ball is11.425 feet, which occurs4.25 feet from the point of release.

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