What is the vertex of $y = x^{2} - 2 x + 1 + {(x - 3)}^{2}$ $y=x^2-2x+1+(x-3)^2$ ?

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Answer 1 · 2018-04-11T17:23:58

$(2, 2)$ $(2,2)$

Let's simplify the expression,

$y = x^{2} - 2 x + 1 + x^{2} + 9 - 6 x$ $" "y=x^2-2x+1+x^2+9-6x$

$\Rightarrow y = 2 x^{2} - 8 x + 10$ $=>" "y=2x^2-8x+10$

$\Rightarrow \frac{y}{2} - 1 = x^{2} - 4 x + 4$ $=>" "y/2-1=x^2-4x+4$

$\Rightarrow \frac{1}{2} (y - 2) = {(x - 2)}^{2}$ $=>" "1/2(y-2)=(x-2)^2$

This is the equation of standard parabola of the form $x^{2} = 4 a y$ $x^2=4ay$

The origin is shifted and so the new vertex is $(2, 2)$ $(2,2)$

What is the vertex of y=x^2-2x+1+(x-3)^2 y=x2−2x+1+(x−3)2y=x^2-2x+1+(x-3)^2 ?