QuestionA ball is thrown from 8 feet high with height $f(x)=-0.2 x^{2}+1.4 x+8$ . Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the equation $f(x)=-0. x^{}+1.4 x+8$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown.

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.2 x^{2}+1.4 x+8$ . 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

Substitute $a=-0.2$ and $b=1.$ into the formula $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{1.}{2(-0.2)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.4}{-0.4} =3.$

STEP 5

The x-coordinate of the vertex is the horizontal distance from where the ball was thrown when it reaches its maximum height. So, the ball reaches its maximum height3.5 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute $x=3.5$ into the equation $f(x)=-0.2 x^{2}+1.4 x+8$ .
$f(3.5) = -0.2 (3.5)^{2}+1.4 (3.5)+8$

STEP 7

Calculate the maximum height of the ball.
$f(3.5) = -0.2 (12.25)+4.9+ = -2.45+4.9+ =10.45$ The maximum height of the ball is10.45 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 8 feet high with height $f(x)=-0.2 x^{2}+1.4 x+8$ . Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the equation $f(x)=-0. x^{}+1.4 x+8$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown.

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.2 x^{2}+1.4 x+8$ . 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

Substitute $a=-0.2$ and $b=1.$ into the formula $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{1.}{2(-0.2)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.4}{-0.4} =3.$

STEP 5

The x-coordinate of the vertex is the horizontal distance from where the ball was thrown when it reaches its maximum height. So, the ball reaches its maximum height3.5 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute $x=3.5$ into the equation $f(x)=-0.2 x^{2}+1.4 x+8$ .
$f(3.5) = -0.2 (3.5)^{2}+1.4 (3.5)+8$

STEP 7

Calculate the maximum height of the ball.
$f(3.5) = -0.2 (12.25)+4.9+ = -2.45+4.9+ =10.45$ The maximum height of the ball is10.45 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 8 feet high with height f(x)=−0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8f(x)=−0.2x2+1.4x+8. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x)f(x), is modeled by the equation f(x)=−0.x+1.4x+8f(x)=-0. x^{}+1.4 x+8f(x)=−0.x+1.4x+8 . xxx is the ball's horizontal distance, in feet, from where it was thrown.

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation f(x)=−0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8f(x)=−0.2x2+1.4x+8. The x-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

Substitute a=−0.2a=-0.2a=−0.2 and b=1.b=1.b=1. into the formula −b2a-\frac{b}{2a}−2ab​ to find the x-coordinate of the vertex.x=−1.2(−0.2)x = -\frac{1.}{2(-0.2)}x=−2(−0.2)1.​

STEP 4

Calculate the x-coordinate of the vertex.x=−1.4−0.4=3.x = -\frac{1.4}{-0.4} =3.x=−−0.41.4​=3.

STEP 5

The x-coordinate of the vertex is the horizontal distance from where the ball was thrown when it reaches its maximum height. So, the ball reaches its maximum height3.5 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute x=3.5x=3.5x=3.5 into the equation f(x)=−0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8f(x)=−0.2x2+1.4x+8.f(3.5)=−0.2(3.5)2+1.4(3.5)+8f(3.5) = -0.2 (3.5)^{2}+1.4 (3.5)+8f(3.5)=−0.2(3.5)2+1.4(3.5)+8

STEP 7

Calculate the maximum height of the ball.f(3.5)=−0.2(12.25)+4.9+=−2.45+4.9+=10.45f(3.5) = -0.2 (12.25)+4.9+ = -2.45+4.9+ =10.45f(3.5)=−0.2(12.25)+4.9+=−2.45+4.9+=10.45The maximum height of the ball is10.45 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 8 feet high with height $f(x)=-0.2 x^{2}+1.4 x+8$ . Find the max height and distance from release.

Assumptions1. The height of the ball, $f(x)$ , is modeled by the equation $f(x)=-0. x^{}+1.4 x+8$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown.

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.2 x^{2}+1.4 x+8$ . The x-coordinate of the vertex of a parabola given by $f(x)=ax^{2}+bx+c$ is given by $-\frac{b}{2a}$ .

Substitute $a=-0.2$ and $b=1.$ into the formula $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{1.}{2(-0.2)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{1.4}{-0.4} =3.$

To find the maximum height of the ball, substitute $x=3.5$ into the equation $f(x)=-0.2 x^{2}+1.4 x+8$ .
$f(3.5) = -0.2 (3.5)^{2}+1.4 (3.5)+8$

Calculate the maximum height of the ball.
$f(3.5) = -0.2 (12.25)+4.9+ = -2.45+4.9+ =10.45$ The maximum height of the ball is10.45 feet, which occurs3.5 feet from the point of release.