QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.7 x+6$ . 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 $f(x)=-0.1 x^{}+0.7 x+6$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown3. We need to find the maximum height of the ball and the distance from where it was thrown

STEP 2

The maximum height of the ball is the maximum value of the function $f(x)$ . This occurs at the vertex of the parabola represented by the equation $f(x)=-0.1 x^{2}+0.7 x+6$ . 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

Plug in the values for $a$ and $b$ from the equation $f(x)=-0.1 x^{2}+0.7 x+6$ into the formula for the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.7}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.7}{2(-0.1)} =3.$

STEP 5

The maximum height of the ball occurs at $x=3.5$ . To find the maximum height, plug $x=3.5$ into the equation $f(x)=-0.1 x^{2}+0.7 x+$ .
$f(3.5)=-0.1 (3.5)^{2}+0.7 (3.5)+$

STEP 6

Calculate the maximum height.
$f(3.5)=-0.1 (3.5)^{2}+0. (3.5)+6 =.75$ The maximum height of the ball is.75 feet, which occurs3.5 feet from the point of release.

Was this helpful?

QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.7 x+6$ . 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 $f(x)=-0.1 x^{}+0.7 x+6$ . $x$ is the ball's horizontal distance, in feet, from where it was thrown3. We need to find the maximum height of the ball and the distance from where it was thrown

STEP 2

The maximum height of the ball is the maximum value of the function $f(x)$ . This occurs at the vertex of the parabola represented by the equation $f(x)=-0.1 x^{2}+0.7 x+6$ . 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

Plug in the values for $a$ and $b$ from the equation $f(x)=-0.1 x^{2}+0.7 x+6$ into the formula for the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.7}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.7}{2(-0.1)} =3.$

STEP 5

The maximum height of the ball occurs at $x=3.5$ . To find the maximum height, plug $x=3.5$ into the equation $f(x)=-0.1 x^{2}+0.7 x+$ .
$f(3.5)=-0.1 (3.5)^{2}+0.7 (3.5)+$

STEP 6

Calculate the maximum height.
$f(3.5)=-0.1 (3.5)^{2}+0. (3.5)+6 =.75$ The maximum height of the ball is.75 feet, which occurs3.5 feet from the point of release.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA ball is thrown from 6 feet high, modeled by f(x)=−0.1x2+0.7x+6f(x)=-0.1 x^{2}+0.7 x+6f(x)=−0.1x2+0.7x+6. Find its max height and distance from launch.

Studdy Solution

STEP 1

STEP 2

STEP 3

Plug in the values for aaa and bbb from the equation f(x)=−0.1x2+0.7x+6f(x)=-0.1 x^{2}+0.7 x+6f(x)=−0.1x2+0.7x+6 into the formula for the x-coordinate of the vertex.x=−b2a=−0.72(−0.1)x = -\frac{b}{2a} = -\frac{0.7}{2(-0.1)}x=−2ab​=−2(−0.1)0.7​

STEP 4

Calculate the x-coordinate of the vertex.x=−0.72(−0.1)=3.x = -\frac{0.7}{2(-0.1)} =3.x=−2(−0.1)0.7​=3.

STEP 5

The maximum height of the ball occurs at x=3.5x=3.5x=3.5. To find the maximum height, plug x=3.5x=3.5x=3.5 into the equation f(x)=−0.1x2+0.7x+f(x)=-0.1 x^{2}+0.7 x+f(x)=−0.1x2+0.7x+.f(3.5)=−0.1(3.5)2+0.7(3.5)+f(3.5)=-0.1 (3.5)^{2}+0.7 (3.5)+f(3.5)=−0.1(3.5)2+0.7(3.5)+

STEP 6

Calculate the maximum height.f(3.5)=−0.1(3.5)2+0.(3.5)+6=.75f(3.5)=-0.1 (3.5)^{2}+0. (3.5)+6 =.75f(3.5)=−0.1(3.5)2+0.(3.5)+6=.75The maximum height of the ball is.75 feet, which occurs3.5 feet from the point of release.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.7 x+6$ . Find its max height and distance from launch.

Plug in the values for $a$ and $b$ from the equation $f(x)=-0.1 x^{2}+0.7 x+6$ into the formula for the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.7}{2(-0.1)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{0.7}{2(-0.1)} =3.$

The maximum height of the ball occurs at $x=3.5$ . To find the maximum height, plug $x=3.5$ into the equation $f(x)=-0.1 x^{2}+0.7 x+$ .
$f(3.5)=-0.1 (3.5)^{2}+0.7 (3.5)+$

Calculate the maximum height.
$f(3.5)=-0.1 (3.5)^{2}+0. (3.5)+6 =.75$ The maximum height of the ball is.75 feet, which occurs3.5 feet from the point of release.