# How do find the vertex and axis of symmetry for a quadratic equation y = x^2 - 4x + 1?

Jun 8, 2015

The vertex form for a quadratic equation is
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$ with a vertex at $\left(a , b\right)$

We can convert the given equation: $y = {x}^{2} - 4 x + 1$ into vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = {x}^{2} - 4 x \textcolor{red}{+ 4} + 1 \textcolor{red}{- 4}$

$\textcolor{w h i t e}{\text{XXXX}}$$y = {\left(x - 2\right)}^{2} + \left(- 3\right)$

So the vertex is at $\left(2 , - 3\right)$

This is the equation of a parabola in normal position (vertical axis of symmetry through the vertex), so
the axis of symmetry is $x = 2$

Jun 8, 2015

You can use the 'complete the square' method.

You're looking for ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
Where $2 a x = - 4 x \to a = - 2 \mathmr{and} {a}^{2} = 4$

So we rewrite as:
$y = {x}^{2} - 4 x + 4 - 3 = {\left(x - 2\right)}^{2} - 3$

Now the axis of symmetry is $x = 2$
(remember to reverse the sign with the $x$)
And the apex is $\left(2 , - 3\right)$
graph{x^2-4x+1 [-7.25, 12.75, -4.8, 5.2]}