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

May 6, 2017

Vertex $\left(- 1 , - 3\right)$

#### Explanation:

First distribute: $\text{ } y = {x}^{2} - 2 x + 1 + 4 x - 3$

Add like terms: $\text{ } y = {x}^{2} + 2 x - 2$

This equation is now in $y = A {x}^{2} + B {x}^{+} C = 0$

The vertex is found when $x = - \frac{B}{2 A} = - \frac{2}{2} = - 1$

and $y = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 2 = 1 - 2 - 2 = - 3$

You can also use completing of the square:

$y = \left({x}^{2} + 2 x\right) - 2$

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

$y = {\left(x + 1\right)}^{2} - 2 - {\left(\frac{2}{2}\right)}^{2}$

$y = {\left(x + 1\right)}^{2} - 3$

Standard form $y = {\left(x - h\right)}^{2} - k$, where the vertex is $\left(h , k\right)$

vertex $= \left(- 1 , - 3\right)$