QuestionA ball is thrown from 6 feet high. Its height is modeled by $f(x)=-0.1 x^{2}+0.6 x+6$ . Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.1 x^{}+0.6 x+6$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown3. We need to find the maximum height of the ball and the horizontal distance at which this occurs

STEP 2

The equation given is a quadratic equation of the form $f(x) = ax^2 + bx + c$ , where $a = -0.1$ , $b =0.6$ , and $c =6$ . The maximum height of the ball can be found by finding the vertex of the parabola represented by this equation. The x-coordinate of the vertex can be found using the formula $-\frac{b}{2a}$ .
$x = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.1)}$

STEP 4

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

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 for $f(x)$ .
$f(x) = -0.1 x^{2}+0. x+$

STEP 6

Plug in the value for $x$ to calculate the maximum height.
$f(3) = -0.1 (3)^{2}+0.6 (3)+6$

STEP 7

Calculate the maximum height of the ball.
$f(3) = -0.1 (9)+1.+6 =7.1$ The maximum height of the ball is7.1 feet, which occurs3 feet from the point of release.

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