QuestionAn athlete releases a shot at $40^{\circ}$ with height modeled by $f(x)=-0.01 x^{2}+0.8 x+5.3$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the quadratic function $f(x)=-0.01x^{}+0.8x+5.3$ . . $x$ represents the shot's horizontal distance in feet from its point of release.
3. We want to find the maximum height of the shot and the distance from the point of release where this occurs.

STEP 2

The maximum or minimum of a quadratic function $ax^{2}+bx+c$ is given by the vertex of the parabola. The x-coordinate of the vertex can be found using the formula $-b/(2a)$ .

STEP 3

In our function $f(x)=-0.01x^{2}+0.8x+5.3$ , $a=-0.01$ and $b=0.8$ . We can plug these values into the formula to find the x-coordinate of the vertex.
$x_{vertex} = -b/(2a)$

STEP 4

Substitute the values of $a$ and $b$ into the formula.
$x_{vertex} = -0.8/(2 \times -0.01)$

STEP 5

Calculate the x-coordinate of the vertex.
$x_{vertex} = -0.8/(-0.02) =40$

STEP 6

The x-coordinate of the vertex represents the horizontal distance from the point of release where the maximum height occurs. So, the maximum height occurs40 feet from the point of release.

STEP 7

To find the maximum height, we substitute the x-coordinate of the vertex into the function $f(x)$ .
$f(x_{vertex}) = -0.01x_{vertex}^{2}+0.x_{vertex}+5.3$

STEP 8

Substitute $x_{vertex} =40$ into the function.
$f(40) = -0.01(40)^{2}+0.8(40)+5.3$

STEP 9

Calculate the maximum height.
$f(40) = -.01(160)+32+5.3 =16+32+5.3 =53.3$ The maximum height of the shot is53.3 feet and this occurs40 feet from the point of release.

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