QuestionAn athlete releases a shot modeled by $f(x)=-0.04 x^{2}+2.1 x+5.2$ . Find the maximum height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the function $f(x)=-0.04 x^{}+.1 x+5.$ , where $x$ is the shot's horizontal distance from its point of release and $f(x)$ is the height of the shot. . 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 given function $f(x)=-0.04 x^{2}+2.1 x+5.2$ is a quadratic function of the form $f(x)=ax^2+bx+c$ , where $a<0$ . The maximum value of this function occurs at the vertex of the parabola, which is given by the formula $x=-\frac{b}{2a}$ .

STEP 3

To find the $x$ -coordinate of the vertex, we plug in the coefficients $a$ and $b$ from the given function into the formula.
$x=-\frac{b}{2a}=-\frac{2.1}{2(-0.04)}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x=-\frac{2.1}{2(-0.04)}=26.25$

STEP 5

The $x$ -coordinate of the vertex is the distance from the point of release at which the maximum height occurs. So, the maximum height occurs26.25 feet from the point of release.

STEP 6

To find the maximum height of the shot, we plug in the $x$ -coordinate of the vertex into the given function.
$f(x)=-0.04 x^{2}+2.1 x+5.2$ $f(26.25)=-0.04 (26.25)^{2}+2.1 (26.25)+5.2$

STEP 7

Calculate the maximum height of the shot.
$f(26.25)=-0.04 (26.25)^{2}+2.1 (26.25)+5.2=35.4375$ The maximum height of the shot is35.4375 feet, which occurs26.25 feet from the point of release.

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