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QuestionA ball is thrown from 6 feet high. Its height is modeled by f(x)=−0.3x2+2.1x+6f(x)=-0.3 x^{2}+2.1 x+6. Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x), is modeled by the quadratic equation f(x)=−0.3x+.1x+6f(x)=-0.3 x^{}+.1 x+6. . xx represents the ball's horizontal distance, in feet, from where it was thrown.
3. We are looking for the maximum height of the ball and the horizontal distance at which this occurs.

STEP 2

The maximum or minimum of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula −b/2a-b/2a.

STEP 3

In our equation, a=−0.3a = -0.3 and b=2.1b =2.1. Plug these values into the formula to find the x-coordinate of the vertex.
x=−b2a=−2.12×−0.3x = -\frac{b}{2a} = -\frac{2.1}{2 \times -0.3}

STEP 4

Calculate the x-coordinate of the vertex.
x=−2.12×−0.3=3.x = -\frac{2.1}{2 \times -0.3} =3.

STEP 5

The x-coordinate of the vertex is the horizontal distance from the point of release at which the maximum height occurs. So, the maximum height occurs3.5 feet from the point of release.

STEP 6

To find the maximum height of the ball, plug the x-coordinate of the vertex into the equation for f(x)f(x).
f(3.5)=−0.3×(3.5)2+2.1×3.5+6f(3.5) = -0.3 \times (3.5)^2 +2.1 \times3.5 +6

STEP 7

Calculate the maximum height of the ball.
f(3.5)=−0.3×(3.5)2+2.1×3.5+6=.75f(3.5) = -0.3 \times (3.5)^2 +2.1 \times3.5 +6 =.75The maximum height of the ball is.75 feet, which occurs3.5 feet from the point of release.

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