Math

QuestionA ball is thrown from 8 feet high. Its height is modeled by f(x)=0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8. Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x), is modeled by the equation f(x)=0.x+1.4x+8f(x)=-0. x^{}+1.4 x+8 . The variable 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 occurs at the vertex of the parabola represented by the equation f(x)=0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8. The xx-coordinate of the vertex of a parabola given by the equation f(x)=ax2+bx+cf(x)=ax^{2}+bx+c can be found using the formula b2a-\frac{b}{2a}.

STEP 3

Plug in the values for aa and bb from the equation f(x)=0.2x2+1.x+8f(x)=-0.2 x^{2}+1. x+8 into the formula b2a-\frac{b}{2a} to find the xx-coordinate of the vertex.
x=1.2(0.2)x = -\frac{1.}{2(-0.2)}

STEP 4

Calculate the xx-coordinate of the vertex.
x=1.40.4=3.x = -\frac{1.4}{-0.4} =3.

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 height3.5 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. We can do this by plugging the xx-coordinate of the vertex into the equation f(x)f(x).

STEP 7

Substitute x=3.5x =3.5 into the equation f(x)=0.2x2+1.4x+f(x)=-0.2 x^{2}+1.4 x+.
f(3.5)=0.2(3.5)2+1.4(3.5)+f(3.5) = -0.2 (3.5)^{2}+1.4 (3.5)+

STEP 8

Calculate the value of f(3.5)f(3.5).
f(3.5)=0.2(12.25)+4.+8=.45f(3.5) = -0.2 (12.25)+4.+8 =.45

STEP 9

So, the maximum height of the ball is9.45 feet, which occurs3.5 feet from the point of release.
The maximum height is9.45 feet, which occurs3.5 feet from the point of release.

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