How do you complete the square to find the vertex for #y=x^2-3x+2#?

Answer 1

To complete the square half the coefficient of x, that is #-3/2#, then square it, that is #9/4#. Now add and subtract it to the expression as follows:

y=#x^2 -3x +9/4 -9/4 +2#

=#(x-3/2)^2 -1/4#. That is how 'complete the square' is done.

Vertex is #(3/2, -1/4)#

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Answer 2

Ethan Adkins

To complete the square to find the vertex for the quadratic function y = x^2 - 3x + 2, follow these steps:

Rewrite the quadratic expression by grouping the x-terms: y = (x^2 - 3x) + 2.
To complete the square, take half of the coefficient of x (-3/2) and square it: (-3/2)^2 = 9/4.
Add and subtract the result from step 2 inside the parentheses: y = (x^2 - 3x + 9/4 - 9/4) + 2.
Rewrite the expression, grouping the perfect square trinomial with the constant term: y = (x^2 - 3x + 9/4) - 9/4 + 2.
Simplify: y = (x - 3/2)^2 - 1/4 + 8/4.
Combine like terms: y = (x - 3/2)^2 + 7/4.

Therefore, the vertex of the parabola is at the point (3/2, 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.