How do you find the vertex of $y = x^2 - 4x + 4$ ?

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Answer 1 · 2018-06-10T16:08:36

$(2,0)$

Method 1:
$y=x^2-4x+4$
complete the square to get:
$y=(x-2)^2$

pick out the x value (2), and the y value (o)
to get:
$(2,0)$

method 2:
$y=x^2-4x+4$
$\frac{dy}{dx} = 2x-4$
let $\frac{dy}{dx}=0$ as the vertex is when the gradient of the curve is 0.

to get $x = 2$
substitute that into the original equation to get $(2,0)$

How do you find the vertex of y = x^2 - 4x + 4y = x^2 - 4x + 4?