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QuestionAn athlete releases a shot at 65∘65^{\circ} with height f(x)=−0.03x2+2.1x+6.3f(x)=-0.03 x^{2}+2.1 x+6.3. Find the max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the equation f(x)=−0.03x+.1x+6.3f(x)=-0.03 x^{}+.1 x+6.3 . xx represents the horizontal distance in feet from the point of release3. The maximum height is represented by the vertex of the parabola formed by the equation4. The vertex of a parabola represented by the equation f(x)=ax+bx+cf(x)=ax^+bx+c is given by (−b/a,f(−b/a))(-b/a, f(-b/a))

STEP 2

We need to find the maximum height of the shot, which is the y-coordinate of the vertex of the parabola. The x-coordinate of the vertex is given by −b/2a-b/2a.
xvertex=−b2ax_{vertex} = -\frac{b}{2a}

STEP 3

Now, plug in the given values for aa and bb to calculate the x-coordinate of the vertex.
xvertex=−2.12×−0.03x_{vertex} = -\frac{2.1}{2 \times -0.03}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=−2.12×−0.03=35x_{vertex} = -\frac{2.1}{2 \times -0.03} =35

STEP 5

Now that we have the x-coordinate of the vertex, we can find the y-coordinate, which is the maximum height of the shot. We can do this by plugging the x-coordinate into the equation for f(x)f(x).
f(xvertex)=−0.03xvertex2+2.1xvertex+.3f(x_{vertex}) = -0.03 x_{vertex}^{2}+2.1 x_{vertex}+.3

STEP 6

Plug in the value for xvertexx_{vertex} to calculate the maximum height.
f(xvertex)=−0.03×352+2.1×35+6.3f(x_{vertex}) = -0.03 \times35^{2}+2.1 \times35+6.3

STEP 7

Calculate the maximum height.
f(xvertex)=−0.03×352+2.1×35+6.3=40.3f(x_{vertex}) = -0.03 \times35^{2}+2.1 \times35+6.3 =40.3The maximum height of the shot is40.3 feet, which occurs35 feet from the point of release.

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