QuestionA ball is thrown from 5 ft high. Its height is modeled by $f(x)=-0.1 x^{2}+0.8 x+5$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball can be modeled by the function $f(x)=-0.1 x^{}+0.8 x+5$ . The variable $x$ represents the ball's horizontal distance from where it was thrown3. We are looking for the maximum height of the ball and the distance at which this occurs

STEP 2

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

STEP 3

Plug in the values for $a$ and $b$ from the function $f(x)$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.8}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.8}{2(-0.1)} =4$

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 $f(x)$ .
$f(x) = -0.1(4)^2 +0.8(4) +5$

STEP 6

Calculate the maximum height of the ball.
$f(x) = -0.1(4)^2 +0.8(4) +5 =6.4$ The maximum height of the ball is6.4 feet, which occurs4 feet from the point of release.

Was this helpful?

QuestionA ball is thrown from 5 ft high. Its height is modeled by $f(x)=-0.1 x^{2}+0.8 x+5$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball can be modeled by the function $f(x)=-0.1 x^{}+0.8 x+5$ . The variable $x$ represents the ball's horizontal distance from where it was thrown3. We are looking for the maximum height of the ball and the distance at which this occurs

STEP 2

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

STEP 3

Plug in the values for $a$ and $b$ from the function $f(x)$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.8}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.8}{2(-0.1)} =4$

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 $f(x)$ .
$f(x) = -0.1(4)^2 +0.8(4) +5$

STEP 6

Calculate the maximum height of the ball.
$f(x) = -0.1(4)^2 +0.8(4) +5 =6.4$ The maximum height of the ball is6.4 feet, which occurs4 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 5 ft high. Its height is modeled by f(x)=−0.1x2+0.8x+5f(x)=-0.1 x^{2}+0.8 x+5f(x)=−0.1x2+0.8x+5. Find the max height and distance.

Studdy Solution

STEP 1

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the function f(x)f(x)f(x). The x-coordinate of the vertex of a parabola given by the equation f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c is given by −b2a-\frac{b}{2a}−2ab​.

STEP 3

Plug in the values for aaa and bbb from the function f(x)f(x)f(x) to calculate the x-coordinate of the vertex.x=−b2a=−0.82(−0.1)x = -\frac{b}{2a} = -\frac{0.8}{2(-0.1)}x=−2ab​=−2(−0.1)0.8​

STEP 4

Calculate the x-coordinate of the vertex.x=−0.82(−0.1)=4x = -\frac{0.8}{2(-0.1)} =4x=−2(−0.1)0.8​=4

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)f(x).f(x)=−0.1(4)2+0.8(4)+5f(x) = -0.1(4)^2 +0.8(4) +5f(x)=−0.1(4)2+0.8(4)+5

STEP 6

Calculate the maximum height of the ball.f(x)=−0.1(4)2+0.8(4)+5=6.4f(x) = -0.1(4)^2 +0.8(4) +5 =6.4f(x)=−0.1(4)2+0.8(4)+5=6.4The maximum height of the ball is6.4 feet, which occurs4 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 5 ft high. Its height is modeled by $f(x)=-0.1 x^{2}+0.8 x+5$ . Find the max height and distance.

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

Plug in the values for $a$ and $b$ from the function $f(x)$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.8}{2(-0.1)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{0.8}{2(-0.1)} =4$

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 $f(x)$ .
$f(x) = -0.1(4)^2 +0.8(4) +5$

Calculate the maximum height of the ball.
$f(x) = -0.1(4)^2 +0.8(4) +5 =6.4$ The maximum height of the ball is6.4 feet, which occurs4 feet from the point of release.