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

1 Answer
Jul 19, 2018

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]