Graph f(x)=2x^2-x+1

Problem

ƒ(x)=2*x2−x+1

Solution

1. Identify the type of function and its orientation. This is a quadratic function in the form ƒ(x)=a*x2+b*x+c Since a=2 which is positive, the parabola opens upward.

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

x=(−(−1))/(2*(2))

x=1/4

3. Calculate the y-coordinate of the vertex by evaluating ƒ(1/4)

ƒ(1/4)=2*(1/4)2−1/4+1

ƒ(1/4)=2*(1/16)−1/4+1

ƒ(1/4)=1/8−2/8+8/8

ƒ(1/4)=7/8

The vertex is at (1/4,7/8)

4. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=2*(0)2−0+1

ƒ(0)=1

The y-intercept is (0,1)

5. Determine the x-intercepts by solving 2*x2−x+1=0 using the discriminant D=b2−4*a*c

D=(−1)2−4*(2)*(1)

D=1−8

D=−7

Since the discriminant is negative, there are no real x-intercepts; the graph does not cross the x-axis.

6. Plot additional points to define the shape. For x=1

ƒ(1)=2*(1)2−1+1

ƒ(1)=2

The point is (1,2)

Final Answer

ƒ(x)=2*x2−x+1

The graph is an upward-opening parabola with vertex (1/4,7/8) y-intercept (0,1) and no x-intercepts.

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