QuestionA ball is thrown from 7 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+7$ . Find its max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.x^+1.4x+7$ . $x$ represents the ball's horizontal distance, in feet, from where it 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 given function $f(x)=-0.2x^2+1.4x+7$ is a quadratic function in the form $f(x)=ax^2+bx+c$ , where $a=-0.2$ , $b=1.4$ , and $c=7$ . The maximum value of a quadratic function occurs at the vertex of the parabola. The $x$ -coordinate of the vertex can be found using the formula $x=-\frac{b}{2a}$ .

STEP 3

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

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x=-\frac{1.4}{2(-0.2)}=3.$

STEP 5

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by substituting $x=3.5$ into the equation $f(x)$ .
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+7$

STEP 6

Calculate the maximum height of the ball.
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+=9.75$ The maximum height of the ball is9.75 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 7 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+7$ . Find its max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is given by the equation $f(x)=-0.x^+1.4x+7$ . $x$ represents the ball's horizontal distance, in feet, from where it 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 given function $f(x)=-0.2x^2+1.4x+7$ is a quadratic function in the form $f(x)=ax^2+bx+c$ , where $a=-0.2$ , $b=1.4$ , and $c=7$ . The maximum value of a quadratic function occurs at the vertex of the parabola. The $x$ -coordinate of the vertex can be found using the formula $x=-\frac{b}{2a}$ .

STEP 3

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

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x=-\frac{1.4}{2(-0.2)}=3.$

STEP 5

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by substituting $x=3.5$ into the equation $f(x)$ .
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+7$

STEP 6

Calculate the maximum height of the ball.
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+=9.75$ The maximum height of the ball is9.75 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 7 feet high. Its height is modeled by f(x)=−0.2x2+1.4x+7f(x)=-0.2 x^{2}+1.4 x+7f(x)=−0.2x2+1.4x+7. Find its max height and distance.

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.x=−b2a=−1.2(−0.2)x=-\frac{b}{2a}=-\frac{1.}{2(-0.2)}x=−2ab​=−2(−0.2)1.​

STEP 4

Calculate the xxx-coordinate of the vertex.x=−1.42(−0.2)=3.x=-\frac{1.4}{2(-0.2)}=3.x=−2(−0.2)1.4​=3.

STEP 5

Now that we have the xxx-coordinate of the vertex, we can find the maximum height of the ball by substituting x=3.5x=3.5x=3.5 into the equation f(x)f(x)f(x).f(3.5)=−0.2(3.5)2+1.4(3.5)+7f(3.5)=-0.2(3.5)^2+1.4(3.5)+7f(3.5)=−0.2(3.5)2+1.4(3.5)+7

STEP 6

Calculate the maximum height of the ball.f(3.5)=−0.2(3.5)2+1.4(3.5)+=9.75f(3.5)=-0.2(3.5)^2+1.4(3.5)+=9.75f(3.5)=−0.2(3.5)2+1.4(3.5)+=9.75The maximum height of the ball is9.75 feet, which occurs3.5 feet from the point of release.

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QuestionA ball is thrown from 7 feet high. Its height is modeled by $f(x)=-0.2 x^{2}+1.4 x+7$ . Find its max height and distance.

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

Calculate the $x$ -coordinate of the vertex.
$x=-\frac{1.4}{2(-0.2)}=3.$

Now that we have the $x$ -coordinate of the vertex, we can find the maximum height of the ball by substituting $x=3.5$ into the equation $f(x)$ .
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+7$

Calculate the maximum height of the ball.
$f(3.5)=-0.2(3.5)^2+1.4(3.5)+=9.75$ The maximum height of the ball is9.75 feet, which occurs3.5 feet from the point of release.