Math

QuestionAn athlete releases a shot at 7070^{\circ} with height f(x)=0.05x2+2.7x+6.2f(x)=-0.05 x^{2}+2.7 x+6.2. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The height of the shot put can be modeled by the equation f(x)=0.05x+.7x+6.f(x)=-0.05 x^{}+.7 x+6. . xx represents 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)

STEP 2

The maximum or minimum of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c is given by the vertex of the parabola. The x-coordinate of the vertex can be found using the formula b2a-\frac{b}{2a}.

STEP 3

Now, plug in the given values for aa and bb to calculate the x-coordinate of the vertex.
x=2.72(0.05)x = -\frac{2.7}{2(-0.05)}

STEP 4

Calculate the x-coordinate of the vertex.
x=2.70.1=27x = -\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 plugging this value into the equation for f(x)f(x).
f(x)=0.05x2+2.7x+.2f(x) = -0.05 x^{2}+2.7 x+.2

STEP 6

Plug in the value for xx to calculate the maximum height.
f(27)=0.05(27)2+2.(27)+6.2f(27) = -0.05 (27)^{2}+2. (27)+6.2

STEP 7

Calculate the maximum height.
f(27)=0.05(729)+72.9+6.2=36.65f(27) = -0.05 (729)+72.9+6.2 =36.65The maximum height of the shot is36.65 feet, which occurs27 feet from the point of release.

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