Find the Vertex f(x)=2x^2-x+1

Problem

ƒ(x)=2*x^2−x+1

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x^2+b*x+c where a=2 b=−1 and c=1

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

x=(−(−1))/(2*(2))

x=1/4

3. Substitute the xcoordinate back into the original function to find the ycoordinate, ƒ(1/4)

ƒ(1/4)=2*(1/4)^2−1/4+1

4. Simplify the expression to determine the value of y

ƒ(1/4)=2*(1/16)−1/4+1

ƒ(1/4)=1/8−2/8+8/8

ƒ(1/4)=7/8

Final Answer

Vertex=(1/4,7/8)

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