Question

How do you graph the parabola `y= x^2-4x+4` using vertex, intercepts and additional points?

Answer 1

Vertex: `(2, 0)` , y intercept: `(0,4)` , x intercept:
`(2,0)` , symmetry line : `x=2` , additional points :
`(0,4) , (4,4) and (1.5, 0.25) , (2.5, 0.25)`

Explanation:

`y=x^2-4 x+4 or y=(x-2)^2+0`

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

being vertex , here `h=2 ,k=0,a=1`

Since `a` is positive, parabola opens upward.

Therefore vertex is at `(h,k) or (2, 0)`

Axis of symmetry is `x= h or x = 2 ;`

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

`y=(x-2)^2 or (x-2)^2= 0 or or x=2 or (2,0)` or

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

`y=(x-2)^2 or y= (0-2)^2 or y=4 or (0,4)` . Graph points:

Distance of vertex from directrix is `d = 1/(4|a|)or 1/4`

The length of a parabola's latus rectum is `4d=1`, where "d" is the

distance from the focus to the vertex. Ends of the latus rectum

are `(1.5, 0.25) and (2.5, 0.25)` Additional point is

`(0,4) and (4,4)`

graph{(x-2)^2 [-10, 10, -5, 5]}[Ans]