QuestionAn athlete releases a shot at $65^{\circ}$ with height modeled by $f(x)=-0.04 x^{2}+2.1 x+5.2$ . Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the equation $f(x)=-0.04 x^{}+.1 x+5.$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. The maximum height of the shot put is the maximum value of $f(x)$

STEP 2

The maximum height of the shot put occurs at the vertex of the parabola represented by the equation $f(x)=-0.04 x^{2}+2.1 x+5.2$ . The x-coordinate of the vertex of a parabola given in the form $f(x)=ax^{2}+bx+c$ can be found using the formula $-\frac{b}{2a}$ .
$x_{vertex} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{2 \times -0.04}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{-0.08} =26.25$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height of the shot put by substituting $x_{vertex}$ into the equation for $f(x)$ .
$f(x_{vertex}) = -0.04 (x_{vertex})^{2}+2.1 x_{vertex}+5.2$

STEP 6

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2$

STEP 7

Calculate the maximum height of the shot put.
$f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2 =29.125$ The maximum height of the shot put is29.125 feet and this occurs26.25 feet from the point of release.

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QuestionAn athlete releases a shot at $65^{\circ}$ with height modeled by $f(x)=-0.04 x^{2}+2.1 x+5.2$ . Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the equation $f(x)=-0.04 x^{}+.1 x+5.$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. The maximum height of the shot put is the maximum value of $f(x)$

STEP 2

The maximum height of the shot put occurs at the vertex of the parabola represented by the equation $f(x)=-0.04 x^{2}+2.1 x+5.2$ . The x-coordinate of the vertex of a parabola given in the form $f(x)=ax^{2}+bx+c$ can be found using the formula $-\frac{b}{2a}$ .
$x_{vertex} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{2 \times -0.04}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{-0.08} =26.25$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height of the shot put by substituting $x_{vertex}$ into the equation for $f(x)$ .
$f(x_{vertex}) = -0.04 (x_{vertex})^{2}+2.1 x_{vertex}+5.2$

STEP 6

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2$

STEP 7

Calculate the maximum height of the shot put.
$f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2 =29.125$ The maximum height of the shot put is29.125 feet and this occurs26.25 feet from the point of release.

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QuestionAn athlete releases a shot at 65∘65^{\circ}65∘ with height modeled by f(x)=−0.04x2+2.1x+5.2f(x)=-0.04 x^{2}+2.1 x+5.2f(x)=−0.04x2+2.1x+5.2. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the equation f(x)=−0.04x+.1x+5.f(x)=-0.04 x^{}+.1 x+5.f(x)=−0.04x+.1x+5. . xxx is the shot's horizontal distance, in feet, from its point of release3. The maximum height of the shot put is the maximum value of f(x)f(x)f(x)

STEP 2

STEP 3

Now, plug in the given values for aaa and bbb to calculate the x-coordinate of the vertex.xvertex=−2.12×−0.04x_{vertex} = -\frac{2.1}{2 \times -0.04}xvertex​=−2×−0.042.1​

STEP 4

Calculate the x-coordinate of the vertex.xvertex=−2.1−0.08=26.25x_{vertex} = -\frac{2.1}{-0.08} =26.25xvertex​=−−0.082.1​=26.25

STEP 5

STEP 6

Plug in the value for xvertexx_{vertex}xvertex​ to calculate the maximum height.f(xvertex)=−0.04(26.25)2+2.1×26.25+5.2f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2f(xvertex​)=−0.04(26.25)2+2.1×26.25+5.2

STEP 7

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QuestionAn athlete releases a shot at $65^{\circ}$ with height modeled by $f(x)=-0.04 x^{2}+2.1 x+5.2$ . Find the max height and distance from release.

Assumptions1. The height of the shot put can be modeled by the equation $f(x)=-0.04 x^{}+.1 x+5.$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. The maximum height of the shot put is the maximum value of $f(x)$

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{2 \times -0.04}$

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.1}{-0.08} =26.25$

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.04 (26.25)^{2}+2.1 \times26.25+5.2$