QuestionAn athlete releases a shot at $30^{\circ}$ with height modeled by $f(x)=-0.01 x^{2}+0.6 x+6.4$ . Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the function $f(x)=-0.01 x^{}+0.6 x+6.4$ . . The height of the shot, $f(x)$ , is measured in feet.
3. The horizontal distance from the point of release, $x$ , is also measured in feet.
4. The shot is released at an angle of $30^{\circ}$ .

STEP 2

The maximum height of the shot can be found by finding the vertex of the parabola represented by the function $f(x)$ .
The x-coordinate of the vertex of a parabola given by the equation $f(x) = ax^{2} + bx + c$ is given by $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.01)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.01)} =30$

STEP 5

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs.Now, to find the maximum height, we substitute $x =30$ into the equation $f(x)$ .

STEP 6

Substitute $x =30$ into the equation $f(x)$ .
$f(30) = -0.01(30)^{2} +0.6(30) +6.4$

STEP 7

Calculate the maximum height of the shot.
$f(30) = -0.01(30)^{2} +0.6(30) +6.4 =16$ The maximum height of the shot is16 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot at $30^{\circ}$ with height modeled by $f(x)=-0.01 x^{2}+0.6 x+6.4$ . Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the function $f(x)=-0.01 x^{}+0.6 x+6.4$ . . The height of the shot, $f(x)$ , is measured in feet.
3. The horizontal distance from the point of release, $x$ , is also measured in feet.
4. The shot is released at an angle of $30^{\circ}$ .

STEP 2

The maximum height of the shot can be found by finding the vertex of the parabola represented by the function $f(x)$ .
The x-coordinate of the vertex of a parabola given by the equation $f(x) = ax^{2} + bx + c$ is given by $-\frac{b}{2a}$ .

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.01)}$

STEP 4

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.01)} =30$

STEP 5

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs.Now, to find the maximum height, we substitute $x =30$ into the equation $f(x)$ .

STEP 6

Substitute $x =30$ into the equation $f(x)$ .
$f(30) = -0.01(30)^{2} +0.6(30) +6.4$

STEP 7

Calculate the maximum height of the shot.
$f(30) = -0.01(30)^{2} +0.6(30) +6.4 =16$ The maximum height of the shot is16 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot at 30∘30^{\circ}30∘ with height modeled by f(x)=−0.01x2+0.6x+6.4f(x)=-0.01 x^{2}+0.6 x+6.4f(x)=−0.01x2+0.6x+6.4. Find max height and distance.

Studdy Solution

STEP 1

STEP 2

The maximum height of the shot can be found by finding the vertex of the parabola represented by the function f(x)f(x)f(x).The x-coordinate of the vertex of a parabola given by the equation f(x)=ax2+bx+cf(x) = ax^{2} + bx + cf(x)=ax2+bx+c is given by −b2a-\frac{b}{2a}−2ab​.

STEP 3

Now, plug in the given values for aaa and bbb to calculate the x-coordinate of the vertex.x=−b2a=−0.62(−0.01)x = -\frac{b}{2a} = -\frac{0.6}{2(-0.01)}x=−2ab​=−2(−0.01)0.6​

STEP 4

Calculate the x-coordinate of the vertex.x=−0.62(−0.01)=30x = -\frac{0.6}{2(-0.01)} =30x=−2(−0.01)0.6​=30

STEP 5

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs.Now, to find the maximum height, we substitute x=30x =30x=30 into the equation f(x)f(x)f(x).

STEP 6

Substitute x=30x =30x=30 into the equation f(x)f(x)f(x).f(30)=−0.01(30)2+0.6(30)+6.4f(30) = -0.01(30)^{2} +0.6(30) +6.4f(30)=−0.01(30)2+0.6(30)+6.4

STEP 7

Calculate the maximum height of the shot.f(30)=−0.01(30)2+0.6(30)+6.4=16f(30) = -0.01(30)^{2} +0.6(30) +6.4 =16f(30)=−0.01(30)2+0.6(30)+6.4=16The maximum height of the shot is16 feet, which occurs30 feet from the point of release.

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QuestionAn athlete releases a shot at $30^{\circ}$ with height modeled by $f(x)=-0.01 x^{2}+0.6 x+6.4$ . Find max height and distance.

The maximum height of the shot can be found by finding the vertex of the parabola represented by the function $f(x)$ .
The x-coordinate of the vertex of a parabola given by the equation $f(x) = ax^{2} + bx + c$ is given by $-\frac{b}{2a}$ .

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x = -\frac{b}{2a} = -\frac{0.6}{2(-0.01)}$

Calculate the x-coordinate of the vertex.
$x = -\frac{0.6}{2(-0.01)} =30$

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs.Now, to find the maximum height, we substitute $x =30$ into the equation $f(x)$ .

Substitute $x =30$ into the equation $f(x)$ .
$f(30) = -0.01(30)^{2} +0.6(30) +6.4$

Calculate the maximum height of the shot.
$f(30) = -0.01(30)^{2} +0.6(30) +6.4 =16$ The maximum height of the shot is16 feet, which occurs30 feet from the point of release.