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

2 Answers
Jun 8, 2015

The vertex form for a quadratic equation is
#color(white)("XXXX")##y =m(x-a)^2+b# with a vertex at #(a,b)#

We can convert the given equation: #y = x^2-4x+1# into vertex form:
#color(white)("XXXX")##y= x^2-4xcolor(red)(+4) +1 color(red)(-4)#

#color(white)("XXXX")##y=(x-2)^2+(-3)#

So the vertex is at #(2,-3)#

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 #(x+a)^2=x^2+2ax+a^2#
Where #2ax=-4x->a=-2and a^2=4#

So we rewrite as:
#y=x^2-4x+4-3=(x-2)^2-3#

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