# How do I convert the equation f(x)=x^2-2x-3 to vertex form?

Jul 29, 2015

$\textcolor{red}{f \left(x\right) = {\left(x - 1\right)}^{2} - 4}$

The vertex form of a quadratic is given by y = a(x – h)^2 + k, where ($h , k$) is the vertex.

The "$a$" in the vertex form is the same "$a$" as in $y = a {x}^{2} + b x + c$.

$f \left(x\right) = {x}^{2} - 2 x - 3$

We convert to the "vertex form" by completing the square.

Step 1. Move the constant to the other side.

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

Step 2. Square the coefficient of $x$ and divide by 4.

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

Step 3. Add this value to each side

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

Step 4. Express the right hand side as a square.

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

Step 5. Isolate $f \left(x\right)$.

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

The equation is now in vertex form.

y = a(x – h)^2 + k, where ($h , k$) is the vertex.

$h = 1$ and $k = - 4$, so the vertex is at ($1 , - 4$).