QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+8$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the quadratic function $f(x)=-0.x^{}+1.4x+8$ . . $x$ is the ball's horizontal distance, in feet, from where it was thrown.
3. We are looking for the maximum height of the ball and the distance from the throwing point where this occurs.

STEP 2

The maximum height of the ball corresponds to the vertex of the parabola represented by the quadratic function. The $x$ -coordinate of the vertex can be found using the formula $-\frac{b}{2a}$ , where $a$ and $b$ are the coefficients of $x^{2}$ and $x$ in the quadratic equation, respectively.
$x_{vertex} = -\frac{b}{2a}$

STEP 3

Substitute the values of $a$ and $b$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.}{2(-0.2)}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.4}{-0.4} =3.$

STEP 5

The maximum height of the ball is the $y$ -coordinate of the vertex, which can be found by substituting the $x$ -coordinate of the vertex into the quadratic function.
$f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8$

STEP 6

Calculate the $y$ -coordinate of the vertex, which is the maximum height of the ball.
$f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8 =9.45$ The maximum height of the ball is9.45 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+8$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the quadratic function $f(x)=-0.x^{}+1.4x+8$ . . $x$ is the ball's horizontal distance, in feet, from where it was thrown.
3. We are looking for the maximum height of the ball and the distance from the throwing point where this occurs.

STEP 2

STEP 3

Substitute the values of $a$ and $b$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.}{2(-0.2)}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.4}{-0.4} =3.$

STEP 5

The maximum height of the ball is the $y$ -coordinate of the vertex, which can be found by substituting the $x$ -coordinate of the vertex into the quadratic function.
$f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8$

STEP 6

Calculate the $y$ -coordinate of the vertex, which is the maximum height of the ball.
$f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8 =9.45$ The maximum height of the ball is9.45 feet, which occurs3.5 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)=−0.2x2+1.4x+8f(x)=-0.2 x^{2}+1.4 x+8f(x)=−0.2x2+1.4x+8. Find its max height and distance from release.

Studdy Solution

STEP 1

STEP 2

STEP 3

Substitute the values of aaa and bbb into the formula to find the xxx-coordinate of the vertex.xvertex=−1.2(−0.2)x_{vertex} = -\frac{1.}{2(-0.2)}xvertex​=−2(−0.2)1.​

STEP 4

Calculate the xxx-coordinate of the vertex.xvertex=−1.4−0.4=3.x_{vertex} = -\frac{1.4}{-0.4} =3.xvertex​=−−0.41.4​=3.

STEP 5

The maximum height of the ball is the yyy-coordinate of the vertex, which can be found by substituting the xxx-coordinate of the vertex into the quadratic function.f(xvertex)=−0.2(3.5)2+1.4(3.5)+8f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8f(xvertex​)=−0.2(3.5)2+1.4(3.5)+8

STEP 6

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QuestionA ball is thrown from 8 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+8$ . Find its max height and distance from release.

Substitute the values of $a$ and $b$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.}{2(-0.2)}$

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{1.4}{-0.4} =3.$

The maximum height of the ball is the $y$ -coordinate of the vertex, which can be found by substituting the $x$ -coordinate of the vertex into the quadratic function.
$f(x_{vertex}) = -0.2(3.5)^{2}+1.4(3.5)+8$