QuestionA ball is thrown from 6 feet high. Its height $f(x)=-0.2 x^{2}+1.7 x+6$ . Find the max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0. x^{}+1.7 x+6$ . 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 the point of release where this occurs

STEP 2

The maximum height of the ball is given by the vertex of the parabola represented by the equation $f(x)=-0.2 x^{2}+1.7 x+6$ . 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

Substitute $a=-0.2$ and $b=1.7$ into the formula for the x-coordinate of the vertex to find the horizontal distance from the point of release where the maximum height occurs.
$x = -\frac{b}{2a} = -\frac{1.7}{2(-0.2)}$

STEP 4

Calculate the value of $x$ .
$x = -\frac{1.7}{2(-0.2)} =4.25$

STEP 5

Substitute $x=4.25$ into the equation for $f(x)$ to find the maximum height of the ball.
$f(4.25)=-0.2 (4.25)^{2}+1.7 (4.25)+$

STEP 6

Calculate the value of $f(4.25)$ .
$f(4.25)=-0.2 (4.25)^{2}+1. (4.25)+6 =8.725$ The maximum height of the ball is8. feet (rounded to the nearest tenth), which occurs4.3 feet (rounded to the nearest tenth) from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height $f(x)=-0.2 x^{2}+1.7 x+6$ . Find the max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet.

Studdy Solution

STEP 1

STEP 2

The maximum height of the ball is given by the vertex of the parabola represented by the equation $f(x)=-0.2 x^{2}+1.7 x+6$ . 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

Substitute $a=-0.2$ and $b=1.7$ into the formula for the x-coordinate of the vertex to find the horizontal distance from the point of release where the maximum height occurs.
$x = -\frac{b}{2a} = -\frac{1.7}{2(-0.2)}$

STEP 4

Calculate the value of $x$ .
$x = -\frac{1.7}{2(-0.2)} =4.25$

STEP 5

Substitute $x=4.25$ into the equation for $f(x)$ to find the maximum height of the ball.
$f(4.25)=-0.2 (4.25)^{2}+1.7 (4.25)+$

STEP 6

Calculate the value of $f(4.25)$ .
$f(4.25)=-0.2 (4.25)^{2}+1. (4.25)+6 =8.725$ The maximum height of the ball is8. feet (rounded to the nearest tenth), which occurs4.3 feet (rounded to the nearest tenth) from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height f(x)=−0.2x2+1.7x+6f(x)=-0.2 x^{2}+1.7 x+6f(x)=−0.2x2+1.7x+6. Find the max height and distance from release. Max height: □\square□ feet, distance: □\square□ feet.

Studdy Solution

STEP 1

STEP 2

STEP 3

Substitute a=−0.2a=-0.2a=−0.2 and b=1.7b=1.7b=1.7 into the formula for the x-coordinate of the vertex to find the horizontal distance from the point of release where the maximum height occurs.x=−b2a=−1.72(−0.2)x = -\frac{b}{2a} = -\frac{1.7}{2(-0.2)}x=−2ab​=−2(−0.2)1.7​

STEP 4

Calculate the value of xxx.x=−1.72(−0.2)=4.25x = -\frac{1.7}{2(-0.2)} =4.25x=−2(−0.2)1.7​=4.25

STEP 5

Substitute x=4.25x=4.25x=4.25 into the equation for f(x)f(x)f(x) to find the maximum height of the ball.f(4.25)=−0.2(4.25)2+1.7(4.25)+f(4.25)=-0.2 (4.25)^{2}+1.7 (4.25)+f(4.25)=−0.2(4.25)2+1.7(4.25)+

STEP 6

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QuestionA ball is thrown from 6 feet high. Its height $f(x)=-0.2 x^{2}+1.7 x+6$ . Find the max height and distance from release. Max height: $\square$ feet, distance: $\square$ feet.

Substitute $a=-0.2$ and $b=1.7$ into the formula for the x-coordinate of the vertex to find the horizontal distance from the point of release where the maximum height occurs.
$x = -\frac{b}{2a} = -\frac{1.7}{2(-0.2)}$

Calculate the value of $x$ .
$x = -\frac{1.7}{2(-0.2)} =4.25$

Substitute $x=4.25$ into the equation for $f(x)$ to find the maximum height of the ball.
$f(4.25)=-0.2 (4.25)^{2}+1.7 (4.25)+$