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

Apr 29, 2016

Vertex form$\text{ } y = {\left(x + 0\right)}^{2} - 3$

So the vertex is at $\left(x , y\right) \to \left(0 , - 3\right)$

This is the same as $y = {x}^{2} - 3$

#### Explanation:

There is an inherent $b x$ term within ${\left(x + 1\right)}^{2}$. Normally you would expect all of $b x$ terms to be within the brackets. One is not! Consequently the brackets have to be expanded so that the excluded term of $- 2 x$ can be incorporated with the term (hidden) in the brackets.

Expanding the brackets $y = \left({x}^{2} + 2 x + 1\right) - 2 x - 4$

Combining terms:$\text{ } y = {x}^{2} + 0 x - 3$

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$\textcolor{b l u e}{\text{Determine the vertex form}}$

Standard form:$\text{ "y=ax^2+bx+c" }$ in your case $a = 1$

Vertex form:$\text{ } y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - a {\left(\frac{b}{2 a}\right)}^{2}$

But $\frac{b}{2 a} = 0 \text{ }$ so $\text{ } - a {\left(\frac{b}{2 a}\right)}^{2} = 0$

$y = {\left(x + 0\right)}^{2} - 3 \text{ " ->" } y = {x}^{2} - 3$