QuestionAn athlete releases a shot put at $70^{\circ}$ with height modeled by $f(x)=-0.06 x^{2}+2.7 x+6.4$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot is modeled by the quadratic function $f(x)=-0.06 x^{}+.7 x+6.4$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. We are looking for the maximum height of the shot and the distance from the point of release at which this occurs

STEP 2

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

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06} =22.$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by evaluating the function at this point.
$f(x_{vertex}) = -0.06 x_{vertex}^{2}+2.7 x_{vertex}+.4$

STEP 6

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2. (22.5)+6.4$

STEP 7

Calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2.7 (22.5)+6.4 =36.15$ The maximum height of the shot is36.15 feet and it occurs22.5 feet from the point of release.

Was this helpful?

QuestionAn athlete releases a shot put at $70^{\circ}$ with height modeled by $f(x)=-0.06 x^{2}+2.7 x+6.4$ . Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot is modeled by the quadratic function $f(x)=-0.06 x^{}+.7 x+6.4$ . $x$ is the shot's horizontal distance, in feet, from its point of release3. We are looking for the maximum height of the shot and the distance from the point of release at which this occurs

STEP 2

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

STEP 3

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06}$

STEP 4

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06} =22.$

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by evaluating the function at this point.
$f(x_{vertex}) = -0.06 x_{vertex}^{2}+2.7 x_{vertex}+.4$

STEP 6

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2. (22.5)+6.4$

STEP 7

Calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2.7 (22.5)+6.4 =36.15$ The maximum height of the shot is36.15 feet and it occurs22.5 feet from the point of release.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionAn athlete releases a shot put at 70∘70^{\circ}70∘ with height modeled by f(x)=−0.06x2+2.7x+6.4f(x)=-0.06 x^{2}+2.7 x+6.4f(x)=−0.06x2+2.7x+6.4. Find the max height and distance.

Studdy Solution

STEP 1

STEP 2

The maximum height of the shot is given by the vertex of the parabola represented by the quadratic function. The x-coordinate of the vertex of a parabola given by f(x)=ax2+bx+cf(x)=ax^{2}+bx+cf(x)=ax2+bx+c is given by −b/2a-b/2a−b/2a.xvertex=−b2ax_{vertex} = -\frac{b}{2a}xvertex​=−2ab​

STEP 3

Now, plug in the given values for aaa and bbb to calculate the x-coordinate of the vertex.xvertex=−2.72×−0.06x_{vertex} = -\frac{2.7}{2 \times -0.06}xvertex​=−2×−0.062.7​

STEP 4

Calculate the x-coordinate of the vertex.xvertex=−2.72×−0.06=22.x_{vertex} = -\frac{2.7}{2 \times -0.06} =22.xvertex​=−2×−0.062.7​=22.

STEP 5

Now that we have the x-coordinate of the vertex, we can find the maximum height by evaluating the function at this point.f(xvertex)=−0.06xvertex2+2.7xvertex+.4f(x_{vertex}) = -0.06 x_{vertex}^{2}+2.7 x_{vertex}+.4f(xvertex​)=−0.06xvertex2​+2.7xvertex​+.4

STEP 6

Plug in the value for xvertexx_{vertex}xvertex​ to calculate the maximum height.f(xvertex)=−0.06(22.5)2+2.(22.5)+6.4f(x_{vertex}) = -0.06 (22.5)^{2}+2. (22.5)+6.4f(xvertex​)=−0.06(22.5)2+2.(22.5)+6.4

STEP 7

Calculate the maximum height.f(xvertex)=−0.06(22.5)2+2.7(22.5)+6.4=36.15f(x_{vertex}) = -0.06 (22.5)^{2}+2.7 (22.5)+6.4 =36.15f(xvertex​)=−0.06(22.5)2+2.7(22.5)+6.4=36.15The maximum height of the shot is36.15 feet and it occurs22.5 feet from the point of release.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionAn athlete releases a shot put at $70^{\circ}$ with height modeled by $f(x)=-0.06 x^{2}+2.7 x+6.4$ . Find the max height and distance.

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

Now, plug in the given values for $a$ and $b$ to calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06}$

Calculate the x-coordinate of the vertex.
$x_{vertex} = -\frac{2.7}{2 \times -0.06} =22.$

Now that we have the x-coordinate of the vertex, we can find the maximum height by evaluating the function at this point.
$f(x_{vertex}) = -0.06 x_{vertex}^{2}+2.7 x_{vertex}+.4$

Plug in the value for $x_{vertex}$ to calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2. (22.5)+6.4$

Calculate the maximum height.
$f(x_{vertex}) = -0.06 (22.5)^{2}+2.7 (22.5)+6.4 =36.15$ The maximum height of the shot is36.15 feet and it occurs22.5 feet from the point of release.