How do you find the vertex for $y = x^2 - 2x$ ?

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Answer 1 · 2018-04-03T07:47:10

The vertex is at $(1,-1)$

We can quite easily see where the vertex of the quadratic function is if we write it in vertex form:

$a(x-h)^2+k$ with vertex at $(h,k)$

To complete the square, we need $h$ to be half the $x$ coefficient, so in this case we have $-2 \/ 2=-1$ :

$(x-1)^2+k=x^2-2x$

$x^2-2x+1+k=x^2-2x$

$k=-1$

This means the vertex form of our quadratic function is:

$y=(x-1)^2-1$

And therefore the vertex is at $(1,-1)$

How do you find the vertex for y = x^2 - 2xy = x^2 - 2x?