QuestionAn athlete releases a shot at $35^{\circ}$ . Its height is modeled by $f(x)=-0.01 x^{2}+0.7 x+5.8$ . Find max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The shot's path is modeled by the equation $f(x)=-0.01 x^{}+0.7 x+5.8$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. $f(x)$ is the shot's height, in feet, from the ground

STEP 2

The maximum height of the shot is given by the maximum value of the function $f(x)$ . This occurs at the vertex of the parabola represented by the function. The $x$ -coordinate of the vertex of a parabola given by $f(x)=ax^2+bx+c$ is given by $-\frac{b}{2a}$ .

STEP 3

Substitute $a=-0.01$ and $b=0.7$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{b}{2a} = -\frac{0.7}{2(-0.01)}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{0.7}{-0.02} =35$

STEP 5

Substitute $x=35$ into the function $f(x)$ to find the maximum height of the shot.
$f(35) = -0.01(35)^2 +0.7(35) +5.8$

STEP 6

Calculate the maximum height of the shot.
$f(35) = -0.01(1225) +24.5 +5.8 =12.45$ The maximum height of the shot is12.45 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot at $35^{\circ}$ . Its height is modeled by $f(x)=-0.01 x^{2}+0.7 x+5.8$ . Find max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The shot's path is modeled by the equation $f(x)=-0.01 x^{}+0.7 x+5.8$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. $f(x)$ is the shot's height, in feet, from the ground

STEP 2

The maximum height of the shot is given by the maximum value of the function $f(x)$ . This occurs at the vertex of the parabola represented by the function. The $x$ -coordinate of the vertex of a parabola given by $f(x)=ax^2+bx+c$ is given by $-\frac{b}{2a}$ .

STEP 3

Substitute $a=-0.01$ and $b=0.7$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{b}{2a} = -\frac{0.7}{2(-0.01)}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{0.7}{-0.02} =35$

STEP 5

Substitute $x=35$ into the function $f(x)$ to find the maximum height of the shot.
$f(35) = -0.01(35)^2 +0.7(35) +5.8$

STEP 6

Calculate the maximum height of the shot.
$f(35) = -0.01(1225) +24.5 +5.8 =12.45$ The maximum height of the shot is12.45 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot at 35∘35^{\circ}35∘. Its height is modeled by f(x)=−0.01x2+0.7x+5.8f(x)=-0.01 x^{2}+0.7 x+5.8f(x)=−0.01x2+0.7x+5.8. Find max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The shot's path is modeled by the equation f(x)=−0.01x+0.7x+5.8f(x)=-0.01 x^{}+0.7 x+5.8f(x)=−0.01x+0.7x+5.8 . xxx is the shot's horizontal distance, in feet, from its point of release3. f(x)f(x)f(x) is the shot's height, in feet, from the ground

STEP 2

STEP 3

Substitute a=−0.01a=-0.01a=−0.01 and b=0.7b=0.7b=0.7 into the formula to find the xxx-coordinate of the vertex.xvertex=−b2a=−0.72(−0.01)x_{vertex} = -\frac{b}{2a} = -\frac{0.7}{2(-0.01)}xvertex​=−2ab​=−2(−0.01)0.7​

STEP 4

Calculate the xxx-coordinate of the vertex.xvertex=−0.7−0.02=35x_{vertex} = -\frac{0.7}{-0.02} =35xvertex​=−−0.020.7​=35

STEP 5

Substitute x=35x=35x=35 into the function f(x)f(x)f(x) to find the maximum height of the shot.f(35)=−0.01(35)2+0.7(35)+5.8f(35) = -0.01(35)^2 +0.7(35) +5.8f(35)=−0.01(35)2+0.7(35)+5.8

STEP 6

Calculate the maximum height of the shot.f(35)=−0.01(1225)+24.5+5.8=12.45f(35) = -0.01(1225) +24.5 +5.8 =12.45f(35)=−0.01(1225)+24.5+5.8=12.45The maximum height of the shot is12.45 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot at $35^{\circ}$ . Its height is modeled by $f(x)=-0.01 x^{2}+0.7 x+5.8$ . Find max height and distance from release.

Assumptions1. The shot's path is modeled by the equation $f(x)=-0.01 x^{}+0.7 x+5.8$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. $f(x)$ is the shot's height, in feet, from the ground

Substitute $a=-0.01$ and $b=0.7$ into the formula to find the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{b}{2a} = -\frac{0.7}{2(-0.01)}$

Calculate the $x$ -coordinate of the vertex.
$x_{vertex} = -\frac{0.7}{-0.02} =35$

Substitute $x=35$ into the function $f(x)$ to find the maximum height of the shot.
$f(35) = -0.01(35)^2 +0.7(35) +5.8$

Calculate the maximum height of the shot.
$f(35) = -0.01(1225) +24.5 +5.8 =12.45$ The maximum height of the shot is12.45 feet, which occurs35 feet from the point of release.