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QuestionA ball is thrown from 5 feet high. Its height is modeled by f(x)=−0.2x2+2.1x+5f(x)=-0.2 x^{2}+2.1 x+5. Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x), can be modeled by the equation f(x)=−0.x+.1x+5f(x)=-0. x^{}+.1 x+5 . xx is 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)=−0.2x2+2.1x+5f(x)=-0.2 x^{2}+2.1 x+5. The xx-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

Now, plug in the given values for aa and bb to find the xx-coordinate of the vertex.
xvertex=−b2a=−2.12(−0.2)x_{vertex} = -\frac{b}{2a} = -\frac{2.1}{2(-0.2)}

STEP 4

Calculate the xx-coordinate of the vertex.
xvertex=−2.12(−0.2)=.25x_{vertex} = -\frac{2.1}{2(-0.2)} =.25

STEP 5

The xx-coordinate of the vertex represents the horizontal distance from where the ball was thrown when it reaches its maximum height. So, the ball reaches its maximum height5.25 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, we need to find the yy-coordinate of the vertex. This can be found by plugging the xx-coordinate of the vertex into the equation for f(x)f(x).
f(xvertex)=−0.2xvertex2+2.1xvertex+5f(x_{vertex}) = -0.2 x_{vertex}^{2}+2.1 x_{vertex}+5

STEP 7

Now, plug in the value for xvertexx_{vertex} to calculate the maximum height.
f(xvertex)=−0.2(5.25)2+2.1(5.25)+5f(x_{vertex}) = -0.2 (5.25)^{2}+2.1 (5.25)+5

STEP 8

Calculate the maximum height of the ball.
f(xvertex)=−0.2(5.25)2+2.1(5.25)+5=8.3125f(x_{vertex}) = -0.2 (5.25)^{2}+2.1 (5.25)+5 =8.3125The maximum height of the ball is8.3125 feet, which occurs5.25 feet from the point of release.

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