What is the vertex of $y = x^{2} - x - 6$ $y= x^2 - x - 6$ ?

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Answer 1 · 2017-04-29T15:19:00

Vertex is $(0.5, - 6.25)$ $(0.5, -6.25)$

$y = x^{2} - x - 6$ $y = x^2 -x-6$

Find the axis of symmetry first using the formula:

$x = \frac{- b}{2 a} \to x = \frac{- (- 1)}{2 \cdot 1}$ $x = (-b)/(2a)" rarr x = (-(-1))/(2*1)$

$x = \frac{1}{2}$ $x = 1/2$

Then find $y$ $y$ by substituting $x = \frac{1}{2}$ $x = 1/2$

$y = {(\frac{1}{2})}^{2} - \frac{1}{2} - 6$ $y = (1/2)^2 -1/2 -6$

$y = - 6 \frac{1}{4}$ $y= -6 1/4$

Vertex is $(0.5, - 6.25)$ $(0.5, -6.25)$

What is the vertex of y=x2−x−6y= x^2 - x - 6 ?