QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.2 x^{2}+1.2 x+6$ . Find its max height and distance from launch.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the equation $f(x)=-0. x^{}+1. x+6$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown3. We are looking for the maximum height of the ball and the distance from where it was thrown when this maximum height occurs

STEP 2

The maximum height of the ball can be found by finding the vertex of the parabolic function $f(x)=-0.2 x^{2}+1.2 x+6$ . The $x$ -coordinate of the vertex of a parabola given by $f(x)=ax^2+bx+c$ is given by $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the $x$ -coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2 \times -0.2}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.2}{2 \times -0.2} =3$

STEP 5

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by plugging this value into the function $f(x)$ .
$f(x) = -0.2 x^{2}+1.2 x+$

STEP 6

Plug in the value for $x$ to calculate the maximum height.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6$

STEP 7

Calculate the maximum height of the ball.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6 =7.2$ The maximum height of the ball is7.2 feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.2 x^{2}+1.2 x+6$ . Find its max height and distance from launch.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the equation $f(x)=-0. x^{}+1. x+6$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown3. We are looking for the maximum height of the ball and the distance from where it was thrown when this maximum height occurs

STEP 2

The maximum height of the ball can be found by finding the vertex of the parabolic function $f(x)=-0.2 x^{2}+1.2 x+6$ . The $x$ -coordinate of the vertex of a parabola given by $f(x)=ax^2+bx+c$ is given by $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the $x$ -coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2 \times -0.2}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.2}{2 \times -0.2} =3$

STEP 5

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by plugging this value into the function $f(x)$ .
$f(x) = -0.2 x^{2}+1.2 x+$

STEP 6

Plug in the value for $x$ to calculate the maximum height.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6$

STEP 7

Calculate the maximum height of the ball.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6 =7.2$ The maximum height of the ball is7.2 feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by f(x)=−0.2x2+1.2x+6f(x)=-0.2 x^{2}+1.2 x+6f(x)=−0.2x2+1.2x+6. Find its max height and distance from launch.

Studdy Solution

STEP 1

STEP 2

The maximum height of the ball can be found by finding the vertex of the parabolic function f(x)=−0.2x2+1.2x+6f(x)=-0.2 x^{2}+1.2 x+6f(x)=−0.2x2+1.2x+6. The xxx-coordinate of the vertex of a parabola given by f(x)=ax2+bx+cf(x)=ax^2+bx+cf(x)=ax2+bx+c is given by −b2a-\frac{b}{2a}−2ab​.

STEP 3

Now, plug in the given values for aaa and bbb to calculate the xxx-coordinate of the vertex.x=−b2a=−1.22×−0.2x = -\frac{b}{2a} = -\frac{1.2}{2 \times -0.2}x=−2ab​=−2×−0.21.2​

STEP 4

Calculate the xxx-coordinate of the vertex.x=−1.22×−0.2=3x = -\frac{1.2}{2 \times -0.2} =3x=−2×−0.21.2​=3

STEP 5

Now that we have the xxx-coordinate of the vertex, we can find the maximum height of the ball by plugging this value into the function f(x)f(x)f(x).f(x)=−0.2x2+1.2x+f(x) = -0.2 x^{2}+1.2 x+f(x)=−0.2x2+1.2x+

STEP 6

Plug in the value for xxx to calculate the maximum height.f(3)=−0.2×32+1.2×3+6f(3) = -0.2 \times3^{2}+1.2 \times3+6f(3)=−0.2×32+1.2×3+6

STEP 7

Calculate the maximum height of the ball.f(3)=−0.2×32+1.2×3+6=7.2f(3) = -0.2 \times3^{2}+1.2 \times3+6 =7.2f(3)=−0.2×32+1.2×3+6=7.2The maximum height of the ball is7.2 feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.2 x^{2}+1.2 x+6$ . Find its max height and distance from launch.

The maximum height of the ball can be found by finding the vertex of the parabolic function $f(x)=-0.2 x^{2}+1.2 x+6$ . The $x$ -coordinate of the vertex of a parabola given by $f(x)=ax^2+bx+c$ is given by $-\frac{b}{2a}$ .

Now, plug in the given values for $a$ and $b$ to calculate the $x$ -coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2 \times -0.2}$

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.2}{2 \times -0.2} =3$

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by plugging this value into the function $f(x)$ .
$f(x) = -0.2 x^{2}+1.2 x+$

Plug in the value for $x$ to calculate the maximum height.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6$

Calculate the maximum height of the ball.
$f(3) = -0.2 \times3^{2}+1.2 \times3+6 =7.2$ The maximum height of the ball is7.2 feet, which occurs3 feet from the point of release.