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

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

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Answer 1 · 2016-03-15T16:00:04

$(\frac{1}{2}, - \frac{13}{2})$ $(1/2, -13/2)$

Explanation:

Vertex of a parabola in the form $a x^{2} + b x + c$ $ax^2+bx+c$ is given by:
$x = - \frac{b}{2 a}$ $x=-b/(2a)$
Note this is only gives the $x$ $x$ -coordinate; we will have to evaluate this value to get the $y$ $y$ -coordinate.

Our parabola $x^{2} - x - 6$ $x^2-x-6$ has $a = 1$ $a=1$ , $b = - 1$ $b=-1$ , and $c = - 6$ $c=-6$ . Using the vertex formula above, we see:
$x = - \frac{- 1}{2 (1)} = \frac{1}{2}$ $x=-(-1)/(2(1))=1/2$

Evaluating $y$ $y$ at this value:
$y = {(\frac{1}{2})}^{2} - (\frac{1}{2}) - 6 = \frac{1}{4} - \frac{1}{2} - 6 = - \frac{13}{2}$ $y=(1/2)^2-(1/2)-6=1/4-1/2-6=-13/2$

Therefore our vertex occurs at the point $(\frac{1}{2}, - \frac{13}{2})$ $(1/2, -13/2)$ .

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

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Explanation:

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

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