Math

QuestionAn athlete releases a shot at 5050^{\circ} with height modeled by f(x)=0.02x2+1.2x+5.3f(x)=-0.02 x^{2}+1.2 x+5.3. Find max height and distance.

Studdy Solution

STEP 1

Assumptions1. The path of the shot is modeled by the function f(x)=0.02x+1.x+5.3f(x)=-0.02 x^{}+1. x+5.3 . xx 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 where this occurs

STEP 2

The maximum height of the shot occurs at the vertex of the parabola represented by the function f(x)f(x). The x-coordinate of the vertex of a parabola given by the equation f(x)=ax2+bx+cf(x)=ax^{2}+bx+c is given by b2a-\frac{b}{2a}.

STEP 3

Plug in the values for aa and bb from the equation f(x)=0.02x2+1.2x+5.3f(x)=-0.02 x^{2}+1.2 x+5.3 to find the x-coordinate of the vertex.
xvertex=b2a=1.22(0.02)x_{vertex} = -\frac{b}{2a} = -\frac{1.2}{2(-0.02)}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=1.22(0.02)=30x_{vertex} = -\frac{1.2}{2(-0.02)} =30

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(xvertex)=0.02xvertex2+1.2xvertex+5.3f(x_{vertex}) = -0.02 x_{vertex}^{2}+1.2 x_{vertex}+5.3

STEP 6

Plug in the value for xvertexx_{vertex} to find the maximum height.
f(30)=0.02(30)2+1.2(30)+5.3f(30) = -0.02 (30)^{2}+1.2 (30)+5.3

STEP 7

Calculate the maximum height.
f(30)=0.02(30)2+1.2(30)+5.3=23.3f(30) = -0.02 (30)^{2}+1.2 (30)+5.3 =23.3The maximum height of the shot is23.3 feet, which occurs30 feet from the point of release.

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