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QuestionA ball is thrown from 6 feet high. Its height is given by f(x)=−0.1x2+0.8x+6f(x)=-0.1 x^{2}+0.8 x+6. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x), is modeled by the equation f(x)=−0.1x+0.8x+6f(x)=-0.1 x^{}+0.8 x+6 . xx represents the ball's horizontal distance, in feet, from where it was thrown3. We are looking for the maximum height of the ball and the distance from where it was thrown when this occurs

STEP 2

The maximum height of the ball can be found by finding the vertex of the parabola represented by the equation f(x)=−0.1x2+0.8x+6f(x)=-0.1 x^{2}+0.8 x+6. The x-coordinate of the vertex of a parabola given in the form f(x)=ax2+bx+cf(x)=ax^{2}+bx+c is given by −b2a-\frac{b}{2a}.

STEP 3

Let's calculate the x-coordinate of the vertex by plugging the coefficients aa and bb from our equation into the formula.
xvertex=−b2a=−0.82(−0.1)x_{vertex} = -\frac{b}{2a} = -\frac{0.8}{2(-0.1)}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=−0.82(−0.1)=4x_{vertex} = -\frac{0.8}{2(-0.1)} =4

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height of the ball by plugging this value into the equation f(x)f(x).
f(xvertex)=−0.1(4)2+0.8(4)+f(x_{vertex}) = -0.1 (4)^{2}+0.8 (4)+

STEP 6

Calculate the maximum height of the ball.
f(xvertex)=−0.1(4)2+0.8(4)+6=8f(x_{vertex}) = -0.1 (4)^{2}+0.8 (4)+6 =8The maximum height of the ball is8 feet, which occurs4 feet from the point of release.

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