# How do you write y = (x+1)(x-3) in Vertex form and Standard form?

Jul 19, 2018

$y = \left(x + 1\right) \left(x - 3\right)$ in vertex form is $y = {\left(x - 1\right)}^{2} - 4$ and in standard form is $y = {x}^{2} - 2 x - 3$

#### Explanation:

In my opinion, it is easier to go from standard for to vertex form so this is the way I will be doing it.

To solve for standard form:

Multiply out (x+1)(x-3) using FOIL so that:

$y = {x}^{2} - 3 x + x - 3$

Simplify the x terms, which gives you standard form:

$y = {x}^{2} - 2 x - 3$

To solve for vertex form:

$y = {x}^{2} - 2 x - 3$

$y + 3 = {x}^{2} - 2 x$

Create a perfect square on the right hand side of the equation:

$y + 3 + 1 = {x}^{2} - 2 x + 1$

*Make sure to balance the equation by adding on the left side what you added on the right

Then simplify the left side of the equation and factor the right side of the equation

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

Finally, subtract 4 from both sides:

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

The format for vertex form is $y = {\left(x - 1\right)}^{2} - 4$