QuestionA ball is thrown from 6 feet high. Its height is given by $f(x)=-0.4 x^{2}+2.1 x+6$ . Find its max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the quadratic function $f(x)=-0.4 x^{}+.1 x+6$ . . $x$ 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 the function $f(x)$ . In a quadratic function of the form $f(x) = ax^2 + bx + c$ , the maximum or minimum value (vertex) occurs at $x = -\frac{b}{2a}$ .
$x_{max} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.)}$

STEP 4

Calculate the value of $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.4)} =2.625$

STEP 5

The maximum height of the ball is given by $f(x_{max})$ .
$f_{max} = f(x_{max}) = -0.4 x_{max}^{2}+2.1 x_{max}+$

STEP 6

Now, plug in the value of $x_{max}$ to calculate $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6$

STEP 7

Calculate the value of $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6 \approx9.4$ The maximum height is approximately9.4 feet, which occurs2.625 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is given by $f(x)=-0.4 x^{2}+2.1 x+6$ . Find its max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, $f(x)$ , is modeled by the quadratic function $f(x)=-0.4 x^{}+.1 x+6$ . . $x$ 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 the function $f(x)$ . In a quadratic function of the form $f(x) = ax^2 + bx + c$ , the maximum or minimum value (vertex) occurs at $x = -\frac{b}{2a}$ .
$x_{max} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.)}$

STEP 4

Calculate the value of $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.4)} =2.625$

STEP 5

The maximum height of the ball is given by $f(x_{max})$ .
$f_{max} = f(x_{max}) = -0.4 x_{max}^{2}+2.1 x_{max}+$

STEP 6

Now, plug in the value of $x_{max}$ to calculate $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6$

STEP 7

Calculate the value of $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6 \approx9.4$ The maximum height is approximately9.4 feet, which occurs2.625 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is given by f(x)=−0.4x2+2.1x+6f(x)=-0.4 x^{2}+2.1 x+6f(x)=−0.4x2+2.1x+6. Find its max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the ball, f(x)f(x)f(x), is modeled by the quadratic function f(x)=−0.4x+.1x+6f(x)=-0.4 x^{}+.1 x+6f(x)=−0.4x+.1x+6. . xxx is the ball's horizontal distance, in feet, from where it was thrown.

STEP 2

STEP 3

Now, plug in the given values for aaa and bbb to calculate xmaxx_{max}xmax​.xmax=−2.12(−0.)x_{max} = -\frac{2.1}{2(-0.)}xmax​=−2(−0.)2.1​

STEP 4

Calculate the value of xmaxx_{max}xmax​.xmax=−2.12(−0.4)=2.625x_{max} = -\frac{2.1}{2(-0.4)} =2.625xmax​=−2(−0.4)2.1​=2.625

STEP 5

The maximum height of the ball is given by f(xmax)f(x_{max})f(xmax​).fmax=f(xmax)=−0.4xmax2+2.1xmax+f_{max} = f(x_{max}) = -0.4 x_{max}^{2}+2.1 x_{max}+fmax​=f(xmax​)=−0.4xmax2​+2.1xmax​+

STEP 6

Now, plug in the value of xmaxx_{max}xmax​ to calculate fmaxf_{max}fmax​.fmax=−0.4(2.625)2+2.1(2.625)+6f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6fmax​=−0.4(2.625)2+2.1(2.625)+6

STEP 7

Calculate the value of fmaxf_{max}fmax​.fmax=−0.4(2.625)2+2.1(2.625)+6≈9.4f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6 \approx9.4fmax​=−0.4(2.625)2+2.1(2.625)+6≈9.4The maximum height is approximately9.4 feet, which occurs2.625 feet from the point of release.

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QuestionA ball is thrown from 6 feet high. Its height is given by $f(x)=-0.4 x^{2}+2.1 x+6$ . Find its max height and distance.

Assumptions1. The height of the ball, $f(x)$ , is modeled by the quadratic function $f(x)=-0.4 x^{}+.1 x+6$ . . $x$ is the ball's horizontal distance, in feet, from where it was thrown.

Now, plug in the given values for $a$ and $b$ to calculate $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.)}$

Calculate the value of $x_{max}$ .
$x_{max} = -\frac{2.1}{2(-0.4)} =2.625$

The maximum height of the ball is given by $f(x_{max})$ .
$f_{max} = f(x_{max}) = -0.4 x_{max}^{2}+2.1 x_{max}+$

Now, plug in the value of $x_{max}$ to calculate $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6$

Calculate the value of $f_{max}$ .
$f_{max} = -0.4 (2.625)^{2}+2.1 (2.625)+6 \approx9.4$ The maximum height is approximately9.4 feet, which occurs2.625 feet from the point of release.