How do I use the vertex formula to determine the vertex of the graph for #y = x^2 - x + 2#?

Answer 1
#y = x^2-x+2#
We want to convert this into the form #y = m(x-a)^2 +b#
Since the coefficient of #x^2# in the given form is #1# #m=1# and the problem is a "complete the squares" problem
#y=x^2-x +(1/2)^2 + 2 -(1/2)^2#
#y= (x-1/2)^2 + 1 3/4#
and the vertex is at #(-1/2,1 3/4)#
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Answer 2

To determine the vertex of the graph of the quadratic function (y = x^2 - x + 2), you can use the vertex formula, which is (x_v = -\frac{b}{2a}), where (a) is the coefficient of the (x^2) term, and (b) is the coefficient of the (x) term.

For the given function (y = x^2 - x + 2), (a = 1) and (b = -1).

Substituting these values into the vertex formula:

(x_v = -\frac{-1}{2 \times 1} = -\frac{-1}{2} = \frac{1}{2}).

So, the x-coordinate of the vertex is (x_v = \frac{1}{2}).

To find the y-coordinate of the vertex, you can substitute the x-coordinate into the original function:

(y = (\frac{1}{2})^2 - \frac{1}{2} + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} = \frac{7}{4}).

Therefore, the vertex of the graph of (y = x^2 - x + 2) is ((\frac{1}{2}, \frac{7}{4})).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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