QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.6 x+6$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by $-0.1x^ +0.6x +6$ . $x$ is 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 occurs at the vertex of the parabola represented by the equation $f(x) = -0.1x^2 +0.6x +6$ . The x-coordinate of the vertex of a parabola given in the form $f(x) = ax^2 + bx + c$ is given by $-\frac{b}{2a}$ .

STEP 3

Substitute the values of $a$ and $b$ into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.1)} =3$

STEP 5

Substitute $x =3$ into the equation $f(x)$ to find the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.(3) +$

STEP 6

Calculate the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.6(3) +6 =6.$ The maximum height of the ball is6. feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.6 x+6$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by $-0.1x^ +0.6x +6$ . $x$ is 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 occurs at the vertex of the parabola represented by the equation $f(x) = -0.1x^2 +0.6x +6$ . The x-coordinate of the vertex of a parabola given in the form $f(x) = ax^2 + bx + c$ is given by $-\frac{b}{2a}$ .

STEP 3

Substitute the values of $a$ and $b$ into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.1)} =3$

STEP 5

Substitute $x =3$ into the equation $f(x)$ to find the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.(3) +$

STEP 6

Calculate the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.6(3) +6 =6.$ The maximum height of the ball is6. feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by f(x)=−0.1x2+0.6x+6f(x)=-0.1 x^{2}+0.6 x+6f(x)=−0.1x2+0.6x+6. Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x)f(x), is given by −0.1x+0.6x+6-0.1x^ +0.6x +6−0.1x+0.6x+6 . xxx is 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

STEP 3

Substitute the values of aaa and bbb into the formula to find the x-coordinate of the vertex.x=−b2a=−0.62(−0.1)x = -\frac{b}{2a} = -\frac{0.6}{2(-0.1)}x=−2ab​=−2(−0.1)0.6​

STEP 4

Calculate the x-coordinate of the vertex.x=−0.62(−0.1)=3x = -\frac{0.6}{2(-0.1)} =3x=−2(−0.1)0.6​=3

STEP 5

Substitute x=3x =3x=3 into the equation f(x)f(x)f(x) to find the maximum height of the ball.f(3)=−0.1(3)2+0.(3)+f(3) = -0.1(3)^2 +0.(3) +f(3)=−0.1(3)2+0.(3)+

STEP 6

Calculate the maximum height of the ball.f(3)=−0.1(3)2+0.6(3)+6=6.f(3) = -0.1(3)^2 +0.6(3) +6 =6.f(3)=−0.1(3)2+0.6(3)+6=6.The maximum height of the ball is6. feet, which occurs3 feet from the point of release.

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QuestionA ball is thrown from 6 feet high, modeled by $f(x)=-0.1 x^{2}+0.6 x+6$ . Find its max height and distance from release.

Assumptions1. The height of the ball, $f(x)$ , is given by $-0.1x^ +0.6x +6$ . $x$ is 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

Substitute the values of $a$ and $b$ into the formula to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.1)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.1)} =3$

Substitute $x =3$ into the equation $f(x)$ to find the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.(3) +$

Calculate the maximum height of the ball.
$f(3) = -0.1(3)^2 +0.6(3) +6 =6.$ The maximum height of the ball is6. feet, which occurs3 feet from the point of release.