QuestionAn athlete releases a shot modeled by $f(x)=-0.01 x^{2}+0.6 x$ . Find the max height and distance from release point.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the function $f(x)=-0.01 x^{}+0.6 x$ . We are looking for the maximum height of the shot and the distance from the point of release where this occurs3. The maximum height of the shot corresponds to the vertex of the parabola represented by the function $f(x)$

STEP 2

The vertex of a parabola given by the equation $f(x) = ax^2 + bx + c$ is at the point $(h, k)$ , where $h = -\frac{b}{2a}$ and $k = f(h)$ .

STEP 3

Let's find the x-coordinate of the vertex, $h$ , by substituting $a = -0.01$ and $b =0.6$ into the formula $h = -\frac{b}{2a}$ .
$h = -\frac{0.6}{2(-0.01)}$

STEP 4

Calculate the value of $h$ .
$h = -\frac{0.6}{-0.02} =30$

STEP 5

Now, let's find the y-coordinate of the vertex, $k$ , by substituting $h =30$ into the function $f(x)$ .
$k = f(30) = -0.01(30)^2 +0.(30)$

STEP 6

Calculate the value of $k$ .
$k = -0.01(900) +18 = -9 +18 =9$ The maximum height of the shot is9 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot modeled by $f(x)=-0.01 x^{2}+0.6 x$ . Find the max height and distance from release point.

Studdy Solution

STEP 1

STEP 2

The vertex of a parabola given by the equation $f(x) = ax^2 + bx + c$ is at the point $(h, k)$ , where $h = -\frac{b}{2a}$ and $k = f(h)$ .

STEP 3

Let's find the x-coordinate of the vertex, $h$ , by substituting $a = -0.01$ and $b =0.6$ into the formula $h = -\frac{b}{2a}$ .
$h = -\frac{0.6}{2(-0.01)}$

STEP 4

Calculate the value of $h$ .
$h = -\frac{0.6}{-0.02} =30$

STEP 5

Now, let's find the y-coordinate of the vertex, $k$ , by substituting $h =30$ into the function $f(x)$ .
$k = f(30) = -0.01(30)^2 +0.(30)$

STEP 6

Calculate the value of $k$ .
$k = -0.01(900) +18 = -9 +18 =9$ The maximum height of the shot is9 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot modeled by f(x)=−0.01x2+0.6xf(x)=-0.01 x^{2}+0.6 xf(x)=−0.01x2+0.6x. Find the max height and distance from release point.

Studdy Solution

STEP 1

STEP 2

The vertex of a parabola given by the equation f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c is at the point (h,k)(h, k)(h,k), where h=−b2ah = -\frac{b}{2a}h=−2ab​ and k=f(h)k = f(h)k=f(h).

STEP 3

Let's find the x-coordinate of the vertex, hhh, by substituting a=−0.01a = -0.01a=−0.01 and b=0.6b =0.6b=0.6 into the formula h=−b2ah = -\frac{b}{2a}h=−2ab​.h=−0.62(−0.01)h = -\frac{0.6}{2(-0.01)}h=−2(−0.01)0.6​

STEP 4

Calculate the value of hhh.h=−0.6−0.02=30h = -\frac{0.6}{-0.02} =30h=−−0.020.6​=30

STEP 5

Now, let's find the y-coordinate of the vertex, kkk, by substituting h=30h =30h=30 into the function f(x)f(x)f(x).k=f(30)=−0.01(30)2+0.(30)k = f(30) = -0.01(30)^2 +0.(30)k=f(30)=−0.01(30)2+0.(30)

STEP 6

Calculate the value of kkk.k=−0.01(900)+18=−9+18=9k = -0.01(900) +18 = -9 +18 =9k=−0.01(900)+18=−9+18=9The maximum height of the shot is9 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot modeled by $f(x)=-0.01 x^{2}+0.6 x$ . Find the max height and distance from release point.

The vertex of a parabola given by the equation $f(x) = ax^2 + bx + c$ is at the point $(h, k)$ , where $h = -\frac{b}{2a}$ and $k = f(h)$ .

Let's find the x-coordinate of the vertex, $h$ , by substituting $a = -0.01$ and $b =0.6$ into the formula $h = -\frac{b}{2a}$ .
$h = -\frac{0.6}{2(-0.01)}$

Calculate the value of $h$ .
$h = -\frac{0.6}{-0.02} =30$

Now, let's find the y-coordinate of the vertex, $k$ , by substituting $h =30$ into the function $f(x)$ .
$k = f(30) = -0.01(30)^2 +0.(30)$

Calculate the value of $k$ .
$k = -0.01(900) +18 = -9 +18 =9$ The maximum height of the shot is9 feet, which occurs30 feet from the point of release.