Graph y=16x-3x^2

Problem

y=16*x−3*x2

Solution

1. Identify the function type and rewrite it in standard form y=a*x2+b*x+c

y=−3*x2+16*x

2. Determine the concavity by looking at the leading coefficient a=−3 Since a<0 the parabola opens downward.

3. Find the x-coordinate of the vertex using the formula x=(−b)/(2*a)

x=(−16)/(2*(−3))

x=8/3

4. Find the y-coordinate of the vertex by substituting x=8/3 back into the original equation.

y=−3*(8/3)2+16*(8/3)

y=−3*(64/9)+128/3

y=−64/3+128/3

y=64/3

5. Find the x-intercepts by setting y=0 and factoring the expression.

0=x*(16−3*x)

x=0

x=16/3

6. Find the y-intercept by setting x=0

y=16*(0)−3*(0)2

y=0

7. Plot the key points including the vertex (8/3,64/3) and the intercepts (0,0) and (16/3,0) then draw a smooth downward-opening curve through them.

Final Answer

y=−3*x2+16*x

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