Question

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

(one-half, seven-quarters)
(one-half, three-halves)
(1, 1)
(1, 3)

Answer 1

Question: Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

Answer: A or 1/2 , 7/4

Explanation:

did it on an assignment on EDGE

Answer 2

`Vertex = ((1)/(2),(7)/(4))`

Explanation:

Given

`f(x) = x^2 - x +2`

Required

The vertex

We have:

`f(x) = x^2 - x +2`

First, we express the equation as:

`f(x) = a(x - h)^2 +k`

Where

`Vertex = (h,k)`

So, we have:

`f(x) = x^2 - x +2`

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Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

--------------------------------------------

`f(x) = x^2 - x +2`

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

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

Expand

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

Factorize

`f(x) = x(x - (1)/(2))- (1)/(2)(x - (1)/(2))+2 -(1)/(4)`

Factor out x - 1/2

`f(x) = (x - (1)/(2))(x - (1)/(2))+2 -(1)/(4)`

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

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

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

Compare to:
`f(x) = a(x - h)^2 +k`

`h = (1)/(2)`

`k = (7)/(4)`

Hence:

`Vertex = ((1)/(2),(7)/(4))`