QuestionA ball thrown from 5 feet high follows $f(x)=-0.6 x^{2}+2.7 x+5$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.6 x^{}+.7 x+5$ . $x$ represents 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 occurs

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.6 x^{2}+2.7 x+5$ . 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

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{2.7}{2(-0.6)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{2.7}{2(-0.6)} =2.25$

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 equation for $f(x)$ .
$f(x) = -0.(2.25)^{2}+2.7(2.25)+5$

STEP 6

Calculate the maximum height of the ball.
$f(x) = -0.6(2.25)^{2}+2.(2.25)+5 =.875$ The maximum height of the ball is.9 feet (rounded to the nearest tenth), which occurs2.25 feet from the point of release.

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QuestionA ball thrown from 5 feet high follows $f(x)=-0.6 x^{2}+2.7 x+5$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.6 x^{}+.7 x+5$ . $x$ represents 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 occurs

STEP 2

The maximum height of the ball occurs at the vertex of the parabola represented by the equation $f(x)=-0.6 x^{2}+2.7 x+5$ . 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

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{2.7}{2(-0.6)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{2.7}{2(-0.6)} =2.25$

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 equation for $f(x)$ .
$f(x) = -0.(2.25)^{2}+2.7(2.25)+5$

STEP 6

Calculate the maximum height of the ball.
$f(x) = -0.6(2.25)^{2}+2.(2.25)+5 =.875$ The maximum height of the ball is.9 feet (rounded to the nearest tenth), which occurs2.25 feet from the point of release.

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QuestionA ball thrown from 5 feet high follows f(x)=−0.6x2+2.7x+5f(x)=-0.6 x^{2}+2.7 x+5f(x)=−0.6x2+2.7x+5. Find its max height and distance from release.

Studdy Solution

STEP 1

STEP 2

STEP 3

Now, plug in the given values for aaa and bbb to calculate the x-coordinate of the vertex.x=−b2a=−2.72(−0.6)x = -\frac{b}{2a} = -\frac{2.7}{2(-0.6)}x=−2ab​=−2(−0.6)2.7​

STEP 4

Calculate the x-coordinate of the vertex.x=−2.72(−0.6)=2.25x = -\frac{2.7}{2(-0.6)} =2.25x=−2(−0.6)2.7​=2.25

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 equation for f(x)f(x)f(x).f(x)=−0.(2.25)2+2.7(2.25)+5f(x) = -0.(2.25)^{2}+2.7(2.25)+5f(x)=−0.(2.25)2+2.7(2.25)+5

STEP 6

Calculate the maximum height of the ball.f(x)=−0.6(2.25)2+2.(2.25)+5=.875f(x) = -0.6(2.25)^{2}+2.(2.25)+5 =.875f(x)=−0.6(2.25)2+2.(2.25)+5=.875The maximum height of the ball is.9 feet (rounded to the nearest tenth), which occurs2.25 feet from the point of release.

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QuestionA ball thrown from 5 feet high follows $f(x)=-0.6 x^{2}+2.7 x+5$ . Find its max height and distance from release.

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{2.7}{2(-0.6)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{2.7}{2(-0.6)} =2.25$

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 equation for $f(x)$ .
$f(x) = -0.(2.25)^{2}+2.7(2.25)+5$

Calculate the maximum height of the ball.
$f(x) = -0.6(2.25)^{2}+2.(2.25)+5 =.875$ The maximum height of the ball is.9 feet (rounded to the nearest tenth), which occurs2.25 feet from the point of release.