Question

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

Answer 1

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]