Math

QuestionAn athlete releases a shot at 6060^{\circ} with height modeled by f(x)=0.02x2+1.7x+5.7f(x)=-0.02 x^{2}+1.7 x+5.7. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The shot's height, f(x)f(x), is modeled by the equation f(x)=0.02x+1.7x+5.7f(x)=-0.02 x^{}+1.7 x+5.7. . xx represents the shot's horizontal distance, in feet, from its point of release.
3. The maximum height of the shot is asked, and at what horizontal distance from the point of release it occurs.

STEP 2

The maximum height of the shot occurs at the vertex of the parabola represented by the equation f(x)=0.02x2+1.7x+5.7f(x)=-0.02 x^{2}+1.7 x+5.7. To find the xx-coordinate of the vertex, we can use the formula b2a-\frac{b}{2a}, where aa and bb are the coefficients of x2x^2 and xx in the equation, respectively.
xvertex=b2ax_{vertex} = -\frac{b}{2a}

STEP 3

Now, plug in the given values for aa and bb to calculate the xx-coordinate of the vertex.
xvertex=1.72×0.02x_{vertex} = -\frac{1.7}{2 \times -0.02}

STEP 4

Calculate the xx-coordinate of the vertex.
xvertex=1.72×0.02=42.x_{vertex} = -\frac{1.7}{2 \times -0.02} =42.

STEP 5

Now that we have the xx-coordinate of the vertex, we can find the maximum height of the shot by plugging xvertexx_{vertex} into the equation f(x)f(x).
f(xvertex)=0.02xvertex2+1.7xvertex+5.7f(x_{vertex}) = -0.02 x_{vertex}^{2}+1.7 x_{vertex}+5.7

STEP 6

Plug in the value for xvertexx_{vertex} to calculate the maximum height.
f(xvertex)=0.02(42.5)2+1.(42.5)+5.f(x_{vertex}) = -0.02 (42.5)^{2}+1. (42.5)+5.

STEP 7

Calculate the maximum height of the shot.
f(xvertex)=0.02(42.5)2+1.7(42.5)+5.7=45.125f(x_{vertex}) = -0.02 (42.5)^{2}+1.7 (42.5)+5.7 =45.125The maximum height of the shot is45.125 feet, which occurs42.5 feet from the point of release.

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