# How do you find the vertex and intercepts for y = x^2 - 4x + 3?

Dec 21, 2015

vertex at $\left(2 , - 1\right)$
y-intercept: $3$
x-intercepts: $3$ and $1$

#### Explanation:

To find the vertex rewrite the given equation $y = {x}^{2} - 4 x + 3$
into vertex form: $y = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$

Complete the square
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 4 x \textcolor{b l u e}{+ {2}^{2}} + 3 \textcolor{b l u e}{- 4}$
Rewrite as a squared binomial and simplified constant (in vertex form)
$\textcolor{w h i t e}{\text{XXX}} y = 1 {\left(x - 2\right)}^{2} + \left(- 1\right)$
with vertex at $\left(2 , - 1\right)$

The y-intercept is the value of $y$ when $x = 0$

For $y = {x}^{2} - 4 x + 3$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = {9}^{2} - 4 \times \left(0\right) + 3 = 3$

The x-intercepts are the values when $y = 0$

Since $y = {x}^{2} - 4 x + 3$
can be factored as $y = \left(x - 3\right) \left(x - 1\right)$
when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} x = 3$ or $x = 1$