How do you find the vertex of $y = x^{2} - x + 2$ $y = x^2 - x + 2$ ?

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How do you find the vertex of $y = x^{2} - x + 2$ $y = x^2 - x + 2$ ?

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Answer 1 · 2015-12-23T11:50:29

Complete the square to establish vertex form

Explanation:

Using completing the squares with $- \frac{1}{2}$ $-1/2$ (half of $b$ $b$ from $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ ) gives
$y = {(x - \frac{1}{2})}^{2} - \frac{1}{4} + 2$ $y = (x - 1/2)^2 -1/4 +2$
So $y = {(x - \frac{1}{2})}^{2} + \frac{7}{4}$ $y = (x-1/2)^2 +7/4$
Hence the vertex is $(\frac{1}{2}, \frac{7}{4})$ $(1/2,7/4)$

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How do you find the vertex of $y = x^{2} - x + 2$ $y = x^2 - x + 2$ ?

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