QuestionAn athlete releases a shot modeled by $f(x)=-0.03 x^{2}+2.1 x+5.3$ . Find the maximum height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, $f(x)$ , is modeled by the equation $f(x)=-0.03 x^{}+.1 x+5.3$ . . $x$ represents the shot's horizontal distance, in feet, from its point of release.
3. We are looking for the maximum height of the shot and the distance from the point of release at which this occurs.

STEP 2

The equation $f(x)=-0.03 x^{2}+2.1 x+5.$ is a quadratic equation in the form $ax^{2}+bx+c$ , where $a=-0.03$ , $b=2.1$ , and $c=5.$ .

STEP 3

The maximum height of the shot occurs at the vertex of the parabola represented by the quadratic equation. The $x$ -coordinate of the vertex can be found using the formula $-\frac{b}{2a}$ .

STEP 4

Substitute the values of $a$ and $b$ into the formula to find the $x$ -coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{2.1}{2(-0.03)}$

STEP 5

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{-0.06} =35$

STEP 6

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 occurs35 feet from the point of release.

STEP 7

To find the maximum height of the shot, substitute the $x$ -coordinate of the vertex into the equation $f(x)$ .
$f(x) = -0.03 x^{2}+2.1 x+5.3$ $f(35) = -0.03 (35)^{2}+2.1 (35)+5.3$

STEP 8

Calculate the maximum height of the shot.
$f(35) = -0.03 (35)^{2}+2.1 (35)+5.3 =40.3$ The maximum height of the shot is40.3 feet and this occurs35 feet from the point of release.

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