QuestionA ball is thrown from 8 feet high. Model: $f(x)=-0.4 x^{2}+2.1 x+8$ . Find max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet. Round to nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.4 x^{}+.1 x+8$ . The variable $x$ represents the horizontal distance, in feet, from where the ball was thrown3. We are looking for the maximum height of the ball and the distance from where it was thrown when this occurs

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.4 x^{2}+2.1 x+8$ . 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 the values of $a$ and $b$ from the equation $f(x)=-0. x^{2}+2.1 x+8$ into the formula $-\frac{b}{2a}$ to find the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{2 \times -0.}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{-0.8} =2.625$

STEP 5

The $x$ -coordinate of the vertex represents the horizontal distance from where the ball was thrown when it reaches its maximum height. Therefore, the ball reaches its maximum height2.625 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute the $x$ -coordinate of the vertex into the equation $f(x)$ .
$f(2.625) = -0.4 \times (2.625)^{2}+2.1 \times2.625+8$

STEP 7

Calculate the maximum height of the ball.
$f(2.625) = -0.4 \times (2.625)^{2}+2.1 \times2.625+ =10.65625$ The maximum height of the ball is10.65625 feet. However, we are asked to round to the nearest tenth, so the maximum height is approximately10.7 feet.
The maximum height is10.7 feet, which occurs2.625 feet from the point of release.

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QuestionA ball is thrown from 8 feet high. Model: $f(x)=-0.4 x^{2}+2.1 x+8$ . Find max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet. Round to nearest tenth.

Studdy Solution

STEP 1

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.4 x^{2}+2.1 x+8$ . 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 the values of $a$ and $b$ from the equation $f(x)=-0. x^{2}+2.1 x+8$ into the formula $-\frac{b}{2a}$ to find the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{2 \times -0.}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{-0.8} =2.625$

STEP 5

The $x$ -coordinate of the vertex represents the horizontal distance from where the ball was thrown when it reaches its maximum height. Therefore, the ball reaches its maximum height2.625 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute the $x$ -coordinate of the vertex into the equation $f(x)$ .
$f(2.625) = -0.4 \times (2.625)^{2}+2.1 \times2.625+8$

STEP 7

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QuestionA ball is thrown from 8 feet high. Model: f(x)=−0.4x2+2.1x+8f(x)=-0.4 x^{2}+2.1 x+8f(x)=−0.4x2+2.1x+8. Find max height and distance from release. Max height: □\square□ feet, distance: □\square□ feet. Round to nearest tenth.

Studdy Solution

STEP 1

STEP 2

STEP 3

Substitute the values of aaa and bbb from the equation f(x)=−0.x2+2.1x+8f(x)=-0. x^{2}+2.1 x+8f(x)=−0.x2+2.1x+8 into the formula −b2a-\frac{b}{2a}−2ab​ to find the xxx-coordinate of the vertex.x=−2.12×−0.x = -\frac{2.1}{2 \times -0.}x=−2×−0.2.1​

STEP 4

Calculate the xxx-coordinate of the vertex.x=−2.1−0.8=2.625x = -\frac{2.1}{-0.8} =2.625x=−−0.82.1​=2.625

STEP 5

The xxx-coordinate of the vertex represents the horizontal distance from where the ball was thrown when it reaches its maximum height. Therefore, the ball reaches its maximum height2.625 feet from where it was thrown.

STEP 6

To find the maximum height of the ball, substitute the xxx-coordinate of the vertex into the equation f(x)f(x)f(x).f(2.625)=−0.4×(2.625)2+2.1×2.625+8f(2.625) = -0.4 \times (2.625)^{2}+2.1 \times2.625+8f(2.625)=−0.4×(2.625)2+2.1×2.625+8

STEP 7

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QuestionA ball is thrown from 8 feet high. Model: $f(x)=-0.4 x^{2}+2.1 x+8$ . Find max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet. Round to nearest tenth.

Substitute the values of $a$ and $b$ from the equation $f(x)=-0. x^{2}+2.1 x+8$ into the formula $-\frac{b}{2a}$ to find the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{2 \times -0.}$

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{2.1}{-0.8} =2.625$

The $x$ -coordinate of the vertex represents the horizontal distance from where the ball was thrown when it reaches its maximum height. Therefore, the ball reaches its maximum height2.625 feet from where it was thrown.

To find the maximum height of the ball, substitute the $x$ -coordinate of the vertex into the equation $f(x)$ .
$f(2.625) = -0.4 \times (2.625)^{2}+2.1 \times2.625+8$