QuestionAn athlete releases a shot modeled by $f(x)=-0.02 x^{2}+1.4 x+5.3$ . Find its maximum height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the quadratic function $f(x)=-0.02x^{}+1.4x+5.3$ . The variable $x$ represents the shot's horizontal distance in feet from its point of release3. The maximum height of the shot put can be found by finding the vertex of the parabola represented by the quadratic function

STEP 2

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

STEP 3

Substitute the coefficients $a$ and $b$ from the given quadratic function into the vertex formula.
$x = -\frac{1.}{2 \times -0.02}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.4}{-0.04} =35$

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, substitute $x=35$ into the given quadratic function.
$f(35) = -0.02(35)^{2} +1.4(35) +5.3$

STEP 6

Calculate the maximum height.
$f(35) = -0.02(1225) +49 +5.3 =70$ The maximum height of the shot is70 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot modeled by $f(x)=-0.02 x^{2}+1.4 x+5.3$ . Find its maximum height and distance from release.

Studdy Solution

STEP 1

STEP 2

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

STEP 3

Substitute the coefficients $a$ and $b$ from the given quadratic function into the vertex formula.
$x = -\frac{1.}{2 \times -0.02}$

STEP 4

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.4}{-0.04} =35$

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, substitute $x=35$ into the given quadratic function.
$f(35) = -0.02(35)^{2} +1.4(35) +5.3$

STEP 6

Calculate the maximum height.
$f(35) = -0.02(1225) +49 +5.3 =70$ The maximum height of the shot is70 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot modeled by f(x)=−0.02x2+1.4x+5.3f(x)=-0.02 x^{2}+1.4 x+5.3f(x)=−0.02x2+1.4x+5.3. Find its maximum height and distance from release.

Studdy Solution

STEP 1

STEP 2

The vertex of a parabola represented by the quadratic function f(x)=ax2+bx+cf(x)=ax^{2}+bx+cf(x)=ax2+bx+c is given by the formula x=−b2ax=-\frac{b}{2a}x=−2ab​.

STEP 3

Substitute the coefficients aaa and bbb from the given quadratic function into the vertex formula.x=−1.2×−0.02x = -\frac{1.}{2 \times -0.02}x=−2×−0.021.​

STEP 4

Calculate the xxx-coordinate of the vertex.x=−1.4−0.04=35x = -\frac{1.4}{-0.04} =35x=−−0.041.4​=35

STEP 5

STEP 6

Calculate the maximum height.f(35)=−0.02(1225)+49+5.3=70f(35) = -0.02(1225) +49 +5.3 =70f(35)=−0.02(1225)+49+5.3=70The maximum height of the shot is70 feet, which occurs35 feet from the point of release.

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QuestionAn athlete releases a shot modeled by $f(x)=-0.02 x^{2}+1.4 x+5.3$ . Find its maximum height and distance from release.

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

Substitute the coefficients $a$ and $b$ from the given quadratic function into the vertex formula.
$x = -\frac{1.}{2 \times -0.02}$

Calculate the $x$ -coordinate of the vertex.
$x = -\frac{1.4}{-0.04} =35$

Calculate the maximum height.
$f(35) = -0.02(1225) +49 +5.3 =70$ The maximum height of the shot is70 feet, which occurs35 feet from the point of release.