Graphing piecewise functions

Question

How do I graph: 

 

f(x)-= [ (x^2-3x+2) / |x-2| if x is not equal to 4 . 

             0 if x=4 

Answer 1

f(x) = (x²-3x+2)/|x-2|

f(4) = 0

 

We cannot divide by zero so x cannot be 2 ... x≠2

 

We can factor the numerator

x²-3x+2 = (x-2)(x-1)

 

If x>2 then:  |x-2| = x-2 because the difference will be positive

For x>2:  f(x) = (x-2)(x-1)/(x-2) = x-1

 

If x=2 then x-1 would be 2-1= 1

Plot the point (2,1) as an open circle since the point is not included

in the graph.  Pick a number greater than 2 to plot another point

for the line f(x) = x-1...   Let x=5... x-1=4.. the point (5,4) is on

the line.

 

Plot the point (5,4) and draw a ray starting from your open circle

at (2,1) going through the point (5,4) ...

LEAVE AN OPEN CIRCLE AT THE POINT (4,3) ON THE RAY SINCE

FOR THE VALUE X=4 WE WERE TOLD F(4)=0

Place a closed dot at the point (4,0) since this value was given as

a condition of the original problem.

 

You now have a ray beginning at (2,1) at an open circle and moving

up to the right with an open circle at (4,3) and a dot at the point (4,0)

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Now you need to look at what happens when x<2

 

(x-2) will be negative if x<2 so |x-2| = -(x-2)

SO  for x<2

 

f(x) = (x-2)(x-1)/-(x-2) = -(x-1) = 1-x

 

If x=2 on this part of the function 1-x = 1-2 = -1

Draw an open circle at the point (2,-1) indicating that the

point is not included on the graph.

 

Pick an x value less than 2 to plot another point on

f(x) = 1-x ...   For example x=0, f(x) = 1

The point (0,1) is on the graph

 

Draw a ray connecting your open circle at (2,-1) through

the point (0,1).

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Your graph is two rays with a jump discontinuity at x=2.

One ray rises from the point (2,1) to the right with a hole

at (4,3).

One ray rises to the left from point (2,-1)

The point (4,0) has a dot at it indicating f(4) = 0