How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #f(x)=2x^2-4x+1#?

1 Answer
Alan P.
Jun 12, 2015

Rewrite #f(x) = 2x^2 -4x+1# in vertex form to determine vertex is at #(1,-1)# and the axis of symmetry #x=1#.
The #f(x)# intercept is at #f(x) = 1#; the x intercepts are at #x= 2+-sqrt(2)/2#

Explanation:

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

#f(x) = 2x^2-4x+1#
extract #m#
#color(white)("XXXX")##f(x) = 2(x^2-2x) +1#
complete the square
#color(white)("XXXX")##f(x) = 2(x^2-2x+1) +1 -2#
#color(white)("XXXX")##f(x)=2(x-1)^2 +(-1)# with a vertex at #(1,-1)#

The given equation is a parabola in standard orientation, so the axis of symmetry is a vertical line through the vertex; that is #x=1#

The #f(x)# and #x# intercepts are determined by solving the equations for #f(x)# with #x=0# and for #x# with #f(x)=0# respectively. (Use the quadratic formula to determine the #x# solution).

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1607 views around the world
You can reuse this answer
Creative Commons License