# How do I convert the equation f(x)=x^2-4x+3 to vertex form?

Jul 29, 2015

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

#### Explanation:

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 $y = a {x}^{2} + b x + c$.

$f \left(x\right) = {x}^{2} - 4 x + 3$

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

Step 1. Move the constant to the other side.

$f \left(x\right) - 3 = {x}^{2} - 4 x$

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

${\left(- 4\right)}^{2} / 4 = \frac{16}{4} = 4$

Step 3. Add this value to each side

$f \left(x\right) - 3 + 4 = {x}^{2} - 4 x + 4$

Step 4. Combine terms.

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

Step 5. Express the right hand side as a square.

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

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

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

The equation is now in vertex form.

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

$h = 2$ and $k = - 1$, so the vertex is at ($2 , - 1$). 