QuestionAn athlete releases a shot at $70^{\circ}$ , modeled by $f(x)=-0.05 x^{2}+2.7 x+5.7$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, f(x), can be modeled by the equation $f(x)=-0.05 x^{}+.7 x+5.7$ . x represents the shot's horizontal distance, in feet, from its point of release.
3. We are looking for the maximum height of the shot and the distance from the point of release when this occurs.

STEP 2

The given function is a quadratic function of the form $f(x) = ax^2 + bx + c$ , where a, b and c are constants. The maximum or minimum of a quadratic function occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula $-b/2a$ .
$x_{vertex} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for a and b to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.05}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{-0.1} =27$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by substituting this value into the function.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2.7 \times27 +5.7$

STEP 6

Calculate the maximum height.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2. \times27 +5. =39.2$ The maximum height of the shot is39.2 feet, which occurs27 feet from the point of release.

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QuestionAn athlete releases a shot at $70^{\circ}$ , modeled by $f(x)=-0.05 x^{2}+2.7 x+5.7$ . Find its max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, f(x), can be modeled by the equation $f(x)=-0.05 x^{}+.7 x+5.7$ . x represents the shot's horizontal distance, in feet, from its point of release.
3. We are looking for the maximum height of the shot and the distance from the point of release when this occurs.

STEP 2

The given function is a quadratic function of the form $f(x) = ax^2 + bx + c$ , where a, b and c are constants. The maximum or minimum of a quadratic function occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula $-b/2a$ .
$x_{vertex} = -\frac{b}{2a}$

STEP 3

Now, plug in the given values for a and b to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.05}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{-0.1} =27$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by substituting this value into the function.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2.7 \times27 +5.7$

STEP 6

Calculate the maximum height.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2. \times27 +5. =39.2$ The maximum height of the shot is39.2 feet, which occurs27 feet from the point of release.

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QuestionAn athlete releases a shot at 70∘70^{\circ}70∘, modeled by f(x)=−0.05x2+2.7x+5.7f(x)=-0.05 x^{2}+2.7 x+5.7f(x)=−0.05x2+2.7x+5.7. Find its max height and distance from release.

Studdy Solution

STEP 1

STEP 2

STEP 3

Now, plug in the given values for a and b to calculate the x-coordinate of the vertex.xvertex=−2.72×−0.05x_{vertex} = -\frac{2.7}{2 \times -0.05}xvertex​=−2×−0.052.7​

STEP 4

Calculate the x-coordinate of the vertex.xvertex=−2.7−0.1=27x_{vertex} = -\frac{2.7}{-0.1} =27xvertex​=−−0.12.7​=27

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by substituting this value into the function.f(xvertex)=−0.05×(27)2+2.7×27+5.7f(x_{vertex}) = -0.05 \times (27)^{2} +2.7 \times27 +5.7f(xvertex​)=−0.05×(27)2+2.7×27+5.7

STEP 6

Calculate the maximum height.f(xvertex)=−0.05×(27)2+2.×27+5.=39.2f(x_{vertex}) = -0.05 \times (27)^{2} +2. \times27 +5. =39.2f(xvertex​)=−0.05×(27)2+2.×27+5.=39.2The maximum height of the shot is39.2 feet, which occurs27 feet from the point of release.

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QuestionAn athlete releases a shot at $70^{\circ}$ , modeled by $f(x)=-0.05 x^{2}+2.7 x+5.7$ . Find its max height and distance from release.

Now, plug in the given values for a and b to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.05}$

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{-0.1} =27$

Now that we have the x-coordinate of the vertex, we can find the maximum height by substituting this value into the function.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2.7 \times27 +5.7$

Calculate the maximum height.
$f(x_{vertex}) = -0.05 \times (27)^{2} +2. \times27 +5. =39.2$ The maximum height of the shot is39.2 feet, which occurs27 feet from the point of release.