How do you graph the parabola y=x24x+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=x24x+4ory=(x2)2+0

This is vertex form of equation ,y=a(xh)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=horx=2;

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

y=(x2)2or(x2)2=0ororx=2or(2,0) or

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

y=(x2)2ory=(02)2ory=4or(0,4) . Graph points:

Distance of vertex from directrix is d=14|a|or14

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]