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

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

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Answer 1 · 2015-06-08T16:48:13

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

y= $x^{2} - 3 x + \frac{9}{4} - \frac{9}{4} + 2$ $x^2 -3x +9/4 -9/4 +2$

= ${(x - \frac{3}{2})}^{2} - \frac{1}{4}$ $(x-3/2)^2 -1/4$ . That is how 'complete the square' is done.

Vertex is $(\frac{3}{2}, - \frac{1}{4})$ $(3/2, -1/4)$

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

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How do you complete the square to find the vertex for y=x2−3x+2y=x^2-3x+2?

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