f ` (x) = 2x - 1 = 0 for turning point

x = 1/2

f (1/2) = 1/2 - 1/2 + 1 = 1

(1/2 , 1) is the vertex

Given the quadratic f(x) = Ax² + Bx + C,

.....VERTEX. The x-coordinate of the vertex is x = -B/2A

.....and the y-coordinate is at (-B/2A, f(-B/2A))

.....x-INTERCEPT. The x-intercept occurs at y=0

For your problem, f(x)=2x²-x+1 we have

.....A = 2, B = -1, and C = 1

.....VERTEX. x = -B/2A = -(-1)/2(2) = ¼

.....y = f(1/4) = 2(¼)² - (¼) + 1 = ⁷⁄₈

.....so the vertex is at (x,y) = (¼,⁷⁄₈).........ANS

.....x-INTERCEPT occurs at y = 0 but note that

.....the vertex is ABOVE the x-axis and since A = 2 > 0,

.....this quadratic is CONCAVE UP so

.....f(x) = 2x²-x+1 will never cross the x-axis

.....(there are no x-intercepts for this quadratic)..........ANS

i) f'(x) = 4x-1 = 0 for minimum

x = 1/4

y = 1/8-1/4+1 = 7/8

Vertex (1/4,7/8)

ii) No x intercept

f(x) =2*(x^2 - (1/2)x) + 1

(x- 1/4)^2= x^2 - (1/2)x + 1/16

x^2 - (1/2)x = (x-1/4)^2 - 1/16

however we have

f(x) = 2* ( (x-1/4)^2 - 1/16) + 1

f(x) = 2*( x-1/4)^2 -2*1/16 + 1

f(x) = 2*(x -1/4)^2 + 1 - 1/8= 2*(x-1/4)^2 + 7/8

so vertex

is (1/4, 7/8) and this is upward facing parabola

therefore , it has no x-intercept

it starts at x =1/4 and only go upward so it never crosses the x-axis

y-intercept when x = 0

f(x) = 0 -0 + 1 = 1

y-int = 1

Subtracting 1/2 does not help, since you need to subtract 1/8

2*(1/4)^2 = 1/8

2 [ x² - x / 2 + 1/4] - 1 / 2 + 1 = 2 [ x - 1 / 4 ]² + 1 / 2 ===> vertex is ( 1 / 4 , 1 / 2 )...f(x) ≥ 1 / 2 ===> NO x intercepts