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

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Answer 1 · 2015-12-23T01:34:05

$(1, - 3)$ $(1,-3)$

Vertex $= (- \frac{b}{2 a}, f (- \frac{b}{2 a}))$ $= (-b/(2a) , f(-b/(2a)))$

In your case,

$- \frac{b}{2 a} = \frac{- (- 2)}{2} = 1$ $-b/(2a) = (- (-2))/2 = 1$

and

$f (1) = 1^{2} - 2 (1) - 2$ $f(1)=1^2 - 2(1) -2$

$= 1 - 2 - 2$ $=1-2-2$

$= - 3$ $= -3$

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