QuestionA ball is thrown from 6 feet high. Its height is modeled by $f(x)=-0.1 x^{2}+1.2 x+6$ . Find max height and distance from throw point.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.1 x^{}+1. x+6$ . $x$ is the ball's horizontal distance in feet from where it was thrown

STEP 2

The maximum height of the ball is the maximum value of $f(x)$ . This occurs at the vertex of the parabola represented by the equation $f(x)=-0.1 x^{2}+1.2 x+6$ . 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 $f(x)=-0.1 x^{2}+1.2 x+6$ into $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.2}{2(-0.1)} =6$

STEP 5

Substitute $x=$ into the equation $f(x)=-0.1 x^{2}+1.2 x+$ to find the maximum height of the ball.
$f()=-0.1 ()^{2}+1.2 ()+$

STEP 6

Calculate the maximum height of the ball.
$f(6)=-0.1 (6)^{2}+1.2 (6)+6 =9.6$ The maximum height of the ball is9.6 feet, which occurs6 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is modeled by $f(x)=-0.1 x^{2}+1.2 x+6$ . Find max height and distance from throw point.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.1 x^{}+1. x+6$ . $x$ is the ball's horizontal distance in feet from where it was thrown

STEP 2

The maximum height of the ball is the maximum value of $f(x)$ . This occurs at the vertex of the parabola represented by the equation $f(x)=-0.1 x^{2}+1.2 x+6$ . 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 $f(x)=-0.1 x^{2}+1.2 x+6$ into $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2(-0.1)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{1.2}{2(-0.1)} =6$

STEP 5

Substitute $x=$ into the equation $f(x)=-0.1 x^{2}+1.2 x+$ to find the maximum height of the ball.
$f()=-0.1 ()^{2}+1.2 ()+$

STEP 6

Calculate the maximum height of the ball.
$f(6)=-0.1 (6)^{2}+1.2 (6)+6 =9.6$ The maximum height of the ball is9.6 feet, which occurs6 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is modeled by f(x)=−0.1x2+1.2x+6f(x)=-0.1 x^{2}+1.2 x+6f(x)=−0.1x2+1.2x+6. Find max height and distance from throw point.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x)f(x), is given by the equation f(x)=−0.1x+1.x+6f(x)=-0.1 x^{}+1. x+6f(x)=−0.1x+1.x+6 . xxx is the ball's horizontal distance in feet from where it was thrown

STEP 2

STEP 3

Substitute the values of aaa and bbb from the equation f(x)=−0.1x2+1.2x+6f(x)=-0.1 x^{2}+1.2 x+6f(x)=−0.1x2+1.2x+6 into −b2a-\frac{b}{2a}−2ab​ to find the x-coordinate of the vertex.x=−b2a=−1.22(−0.1)x = -\frac{b}{2a} = -\frac{1.2}{2(-0.1)}x=−2ab​=−2(−0.1)1.2​

STEP 4

Calculate the x-coordinate of the vertex.x=−1.22(−0.1)=6x = -\frac{1.2}{2(-0.1)} =6x=−2(−0.1)1.2​=6

STEP 5

Substitute x=x=x= into the equation f(x)=−0.1x2+1.2x+f(x)=-0.1 x^{2}+1.2 x+f(x)=−0.1x2+1.2x+ to find the maximum height of the ball.f()=−0.1()2+1.2()+f()=-0.1 ()^{2}+1.2 ()+f()=−0.1()2+1.2()+

STEP 6

Calculate the maximum height of the ball.f(6)=−0.1(6)2+1.2(6)+6=9.6f(6)=-0.1 (6)^{2}+1.2 (6)+6 =9.6f(6)=−0.1(6)2+1.2(6)+6=9.6The maximum height of the ball is9.6 feet, which occurs6 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is modeled by $f(x)=-0.1 x^{2}+1.2 x+6$ . Find max height and distance from throw point.

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.1 x^{}+1. x+6$ . $x$ is the ball's horizontal distance in feet from where it was thrown

Substitute the values of $a$ and $b$ from the equation $f(x)=-0.1 x^{2}+1.2 x+6$ into $-\frac{b}{2a}$ to find the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{1.2}{2(-0.1)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{1.2}{2(-0.1)} =6$

Substitute $x=$ into the equation $f(x)=-0.1 x^{2}+1.2 x+$ to find the maximum height of the ball.
$f()=-0.1 ()^{2}+1.2 ()+$

Calculate the maximum height of the ball.
$f(6)=-0.1 (6)^{2}+1.2 (6)+6 =9.6$ The maximum height of the ball is9.6 feet, which occurs6 feet from the point of release.