What is the vertex form of $y=x^2 - 2 x - 5$ ?

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Answer 1 · 2018-05-17T20:04:59

Vertex form is $y= (x-1)^2 -6$

The vertex is the point $(1,-6)$

Vertex form is $y = a(x+p)^2 +q$ with the vertex at $(-p, q)$

This is derived by the process of completing the square.

The quadratic trinomial $x^2 -2x-5$ is in the form $ax^2 +bx+c$

$x^2 -2x-5$ is not a perfect square.

We need to add the correct constant which will make it a square.
This is found from $(b/2)^2$ which in this case is $((-2)/2)^2 = color(blue)(1)$

$y = x^2 -2x color(blue)(+1) color(blue)(-1) -5" "larr (color(blue)(+1-1=0))$

$y = (x^2 -2x color(blue)(+1))+ ( color(blue)(-1) -5)$

$y= (x-1)^2 -6" "larr$ this is vertex form.

The vertex is the point $(1,-6)$

What is the vertex form of y=x^2 - 2 x - 5 y=x^2 - 2 x - 5 ?