QuestionAn athlete releases a shot put at $65^{\circ}$ . The height is modeled by $f(x)=-0.03 x^{2}+21 x+5.3$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, $f(x)$ , is modeled by the quadratic function $f(x)=-0.03 x^{}+21 x+5.3$ . . $x$ is the shot's horizontal distance, in feet, from its point of release.
3. We need to find the maximum height of the shot and the distance from the point of release when this occurs.

STEP 2

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

STEP 3

In our case, $a = -0.03$ and $b =21$ . Let's plug these values into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{21}{2(-0.03)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{21}{2(-0.03)} =350$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by plugging this value into the equation for $f(x)$ .
$f(x) = -0.03 x^{2}+21 x+5.3$

STEP 6

Substitute $x =350$ into the equation to find the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3$

STEP 7

Calculate the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3 =3673.3$ The maximum height of the shot is $3673.3$ feet, which occurs $350$ feet from the point of release.

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QuestionAn athlete releases a shot put at $65^{\circ}$ . The height is modeled by $f(x)=-0.03 x^{2}+21 x+5.3$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, $f(x)$ , is modeled by the quadratic function $f(x)=-0.03 x^{}+21 x+5.3$ . . $x$ is the shot's horizontal distance, in feet, from its point of release.
3. We need to find the maximum height of the shot and the distance from the point of release when this occurs.

STEP 2

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

STEP 3

In our case, $a = -0.03$ and $b =21$ . Let's plug these values into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{21}{2(-0.03)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{21}{2(-0.03)} =350$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by plugging this value into the equation for $f(x)$ .
$f(x) = -0.03 x^{2}+21 x+5.3$

STEP 6

Substitute $x =350$ into the equation to find the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3$

STEP 7

Calculate the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3 =3673.3$ The maximum height of the shot is $3673.3$ feet, which occurs $350$ feet from the point of release.

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QuestionAn athlete releases a shot put at 65∘65^{\circ}65∘. The height is modeled by f(x)=−0.03x2+21x+5.3f(x)=-0.03 x^{2}+21 x+5.3f(x)=−0.03x2+21x+5.3. Find the max height and distance.

Studdy Solution

STEP 1

STEP 2

The maximum or minimum of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c is given by the vertex of the parabola. The x-coordinate of the vertex can be found using the formula −b2a-\frac{b}{2a}−2ab​.

STEP 3

In our case, a=−0.03a = -0.03a=−0.03 and b=21b =21b=21. Let's plug these values into the formula to find the x-coordinate of the vertex.x=−b2a=−212(−0.03)x = -\frac{b}{2a} = -\frac{21}{2(-0.03)}x=−2ab​=−2(−0.03)21​

STEP 4

Calculate the x-coordinate of the vertex.x=−212(−0.03)=350x = -\frac{21}{2(-0.03)} =350x=−2(−0.03)21​=350

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by plugging this value into the equation for f(x)f(x)f(x).f(x)=−0.03x2+21x+5.3f(x) = -0.03 x^{2}+21 x+5.3f(x)=−0.03x2+21x+5.3

STEP 6

Substitute x=350x =350x=350 into the equation to find the maximum height.f(350)=−0.03(350)2+21(350)+5.3f(350) = -0.03 (350)^{2}+21 (350)+5.3f(350)=−0.03(350)2+21(350)+5.3

STEP 7

Calculate the maximum height.f(350)=−0.03(350)2+21(350)+5.3=3673.3f(350) = -0.03 (350)^{2}+21 (350)+5.3 =3673.3f(350)=−0.03(350)2+21(350)+5.3=3673.3The maximum height of the shot is 3673.33673.33673.3 feet, which occurs 350350350 feet from the point of release.

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QuestionAn athlete releases a shot put at $65^{\circ}$ . The height is modeled by $f(x)=-0.03 x^{2}+21 x+5.3$ . Find the max height and distance.

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

In our case, $a = -0.03$ and $b =21$ . Let's plug these values into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{21}{2(-0.03)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{21}{2(-0.03)} =350$

Now that we have the x-coordinate of the vertex, we can find the maximum height by plugging this value into the equation for $f(x)$ .
$f(x) = -0.03 x^{2}+21 x+5.3$

Substitute $x =350$ into the equation to find the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3$

Calculate the maximum height.
$f(350) = -0.03 (350)^{2}+21 (350)+5.3 =3673.3$ The maximum height of the shot is $3673.3$ feet, which occurs $350$ feet from the point of release.