How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. #y=x^2-2x+1#?

1 Answer
Binayaka C.
Aug 4, 2018

Vertex is at # (1, 0)# ,axis of symmetry is # x = 1 #
x intercept is #x=1# ,y intercept is #y=1#, additional
points on parabola are #(-1,4),(3,4)#

Explanation:

#y=x^2-2 x+1 or y=(x-1)^2+0#

This is vertex form of equation ,#y=a(x-h)^2+k ; (h,k)#

being vertex , here #h=1 ,k=0,a=1 #

Since #a# is positive, parabola opens upward.

Therefore vertex is at #(h,k) or (1, 0)#

Axis of symmetry is #x= h or x = 1 #

x-intercept is found by putting #y=0# in the equation

#y=(x-1)^2 or (x-1)^2= 0 :.x=1 or (1,0) # or

x intercept is at #x=1 or(1,0)#

y-intercept is found by putting #x=0# in the equation

#y=(x-1)^2 or y= (0-1)^2 :. y=1 or (0,1)#

y intercept is at #y=1 or(0,1)#

Additional points: Let #x=3 :. y= (3-1)^2=4 or (3,4) # and

#x=-1 :. y= (-1-1)^2=4 or (-1,4) #

graph{x^2-2 x +1 [-10, 10, -5, 5]} [Ans]

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
4668 views around the world
You can reuse this answer
Creative Commons License