What is the the vertex of $y = (x-1)^2+4x-3$ ?

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Answer 1 · 2017-05-06T16:31:53

Vertex $(-1, -3)$

First distribute: $" "y = x^2 - 2x + 1 + 4x -3$

Add like terms: $" "y = x^2 +2x -2$

This equation is now in $y = Ax^2 + Bx^+C = 0$

The vertex is found when $x = -B/(2A) = -2/2 = -1$

and $y = (-1)^2 + 2(-1) - 2 = 1 -2 - 2 = -3$

You can also use completing of the square:

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

Half the x-term and complete the square by subtracting the square of that value:

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

$y = (x+1)^2 - 3$

Standard form $y = (x-h)^2 -k$ , where the vertex is $(h, k)$

vertex $= (-1, -3)$

What is the the vertex of y = (x-1)^2+4x-3y = (x-1)^2+4x-3?