QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.3 x^{2}+1.7 x+8$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball is modeled by the equation $f(x)=-0.3 x^{}+1.7 x+8$ . $x$ 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 at which this occurs

STEP 2

The given function $f(x)=-0. x^{2}+1.7 x+8$ is a quadratic function, and its graph is a parabola. The maximum or minimum value of a parabola occurs at its vertex. The x-coordinate of the vertex of a parabola given by $f(x) = ax^2 + bx + c$ is $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{2 \times -0.3}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{-0.6} =2.83 \, feet$

STEP 5

Now, we can find the maximum height of the ball by substituting the x-coordinate of the vertex into the equation $f(x)$ .
$f(2.83) = -0.3 \times (2.83)^{2} +1.7 \times2.83 +8$

STEP 6

Calculate the maximum height of the ball.
$f(2.83) = -0.3 \times (2.83)^{2} +1. \times2.83 +8 =10.8 \, feet$ The maximum height of the ball is10.8 feet, which occurs2.83 feet from the point of release.

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QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.3 x^{2}+1.7 x+8$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball is modeled by the equation $f(x)=-0.3 x^{}+1.7 x+8$ . $x$ 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 at which this occurs

STEP 2

The given function $f(x)=-0. x^{2}+1.7 x+8$ is a quadratic function, and its graph is a parabola. The maximum or minimum value of a parabola occurs at its vertex. The x-coordinate of the vertex of a parabola given by $f(x) = ax^2 + bx + c$ is $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{2 \times -0.3}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{-0.6} =2.83 \, feet$

STEP 5

Now, we can find the maximum height of the ball by substituting the x-coordinate of the vertex into the equation $f(x)$ .
$f(2.83) = -0.3 \times (2.83)^{2} +1.7 \times2.83 +8$

STEP 6

Calculate the maximum height of the ball.
$f(2.83) = -0.3 \times (2.83)^{2} +1. \times2.83 +8 =10.8 \, feet$ The maximum height of the ball is10.8 feet, which occurs2.83 feet from the point of release.

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QuestionA ball is thrown from 8 feet high. Its height is modeled by f(x)=−0.3x2+1.7x+8f(x)=-0.3 x^{2}+1.7 x+8f(x)=−0.3x2+1.7x+8. Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball is modeled by the equation f(x)=−0.3x+1.7x+8f(x)=-0.3 x^{}+1.7 x+8f(x)=−0.3x+1.7x+8 . xxx 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 at which this occurs

STEP 2

STEP 3

Now, plug in the given values for aaa and bbb to calculate the x-coordinate of the vertex.x=−1.72×−0.3x = -\frac{1.7}{2 \times -0.3}x=−2×−0.31.7​

STEP 4

Calculate the x-coordinate of the vertex.x=−1.7−0.6=2.83 feetx = -\frac{1.7}{-0.6} =2.83 \, feetx=−−0.61.7​=2.83feet

STEP 5

Now, we can find the maximum height of the ball by substituting the x-coordinate of the vertex into the equation f(x)f(x)f(x).f(2.83)=−0.3×(2.83)2+1.7×2.83+8f(2.83) = -0.3 \times (2.83)^{2} +1.7 \times2.83 +8f(2.83)=−0.3×(2.83)2+1.7×2.83+8

STEP 6

Calculate the maximum height of the ball.f(2.83)=−0.3×(2.83)2+1.×2.83+8=10.8 feetf(2.83) = -0.3 \times (2.83)^{2} +1. \times2.83 +8 =10.8 \, feetf(2.83)=−0.3×(2.83)2+1.×2.83+8=10.8feetThe maximum height of the ball is10.8 feet, which occurs2.83 feet from the point of release.

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QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.3 x^{2}+1.7 x+8$ . Find its max height and distance from release.

Assumptions1. The height of the ball is modeled by the equation $f(x)=-0.3 x^{}+1.7 x+8$ . $x$ 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 at which this occurs

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{2 \times -0.3}$

Calculate the x-coordinate of the vertex.
$x = -\frac{1.7}{-0.6} =2.83 \, feet$

Now, we can find the maximum height of the ball by substituting the x-coordinate of the vertex into the equation $f(x)$ .
$f(2.83) = -0.3 \times (2.83)^{2} +1.7 \times2.83 +8$

Calculate the maximum height of the ball.
$f(2.83) = -0.3 \times (2.83)^{2} +1. \times2.83 +8 =10.8 \, feet$ The maximum height of the ball is10.8 feet, which occurs2.83 feet from the point of release.