Math

QuestionA ball is thrown from 8 feet high. Its height is modeled by f(x)=0.1x2+0.6x+8f(x)=-0.1 x^{2}+0.6 x+8. Find its max height and distance from launch.

Studdy Solution

STEP 1

Assumptions1. The height of the ball can be modeled by the quadratic function f(x)=0.1x+0.6x+8f(x)=-0.1 x^{}+0.6 x+8. . The variable xx represents the horizontal distance, in feet, from where the ball was thrown.
3. We are looking for the maximum height of the ball and the distance from where it was thrown when this maximum height occurs.

STEP 2

The maximum height of the ball occurs at the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula b2a-\frac{b}{2a}, where aa and bb are the coefficients of the quadratic and linear terms, respectively.
xvertex=b2ax_{vertex} = -\frac{b}{2a}

STEP 3

Substitute the values a=0.1a=-0.1 and b=0.6b=0.6 into the formula.
xvertex=0.62(0.1)x_{vertex} = -\frac{0.6}{2(-0.1)}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=0.60.2=3x_{vertex} = -\frac{0.6}{-0.2} =3

STEP 5

The maximum height of the ball is the y-coordinate of the vertex, which can be found by substituting the x-coordinate of the vertex into the function f(x)f(x).
f(xvertex)=0.1(3)2+0.(3)+8f(x_{vertex}) = -0.1 (3)^{2}+0. (3)+8

STEP 6

Calculate the maximum height of the ball.
f(xvertex)=0.1(9)+1.8+8=.1f(x_{vertex}) = -0.1 (9)+1.8+8 =.1The maximum height of the ball is.1 feet, which occurs3 feet from the point of release.

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