Find the Vertex y=2x^2-8x+6

Problem

y=2*x2−8*x+6

Solution

1. Identify the coefficients of the quadratic equation in the form y=a*x2+b*x+c

a=2

b=−8

c=6

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

x=(−(−8))/(2*(2))

x=8/4

x=2

3. Substitute the x-coordinate back into the original equation to find the y-coordinate.

y=2*(2)2−8*(2)+6

y=2*(4)−16+6

y=8−16+6

y=−2

4. State the vertex as an ordered pair (x,y)

Vertex=(2,−2)

Final Answer

Vertex of *y=2*x2−8*x+6=(2,−2)

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