How do you find the vertex of $y = x^{2} - 5 x + 4$ $y=x^2 - 5x + 4$ ?

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Answer 1 · 2018-06-06T21:07:06

$(\frac{5}{2}, \frac{- 9}{4})$ $(5/2,(-9)/4)$

$\frac{- b}{2 a}$ $(-b)/(2a)$ gives the x coord of the vertex

$\Rightarrow \frac{- - 5}{2} = 2 \frac{1}{2}$ $=> (- -5)/2=2 1/2$

When $x = 2 \frac{1}{2}, y = \frac{25}{4} - \frac{25}{2} + 4 = \frac{- 9}{4}$ $x=2 1/2, y=25/4-25/2+4=(-9)/4$

How do you find the vertex of y=x2−5x+4y=x^2 - 5x + 4?