Transcript graphing-piecewise-functions-1

WARM-UP
1. Is the ordered pair (-6,-5) a solution to the
inequality 2x -4y < 8?
2. Solve the linear inequalities:
a. y > (3/5) x – 3
b. 4x – 2y > -3
Graphing Piecewise Functions
• Up to now, we’ve been looking at
functions represented by a single
equation.
• In real life, however, functions are
represented by a combination of
equations, each corresponding to
a part of the domain.
• These are called piecewise
functions.
Evaluate f(x) when x=0, x=2, x=4
 x  2, if x  2
f ( x)  
 2 x  1, if x  2
•First you have to figure out which equation to use
•You NEVER use both
X=0
So:
This one fits
Into the top
0+2=2
equation
f(0)=2
X=2
This
So: one fits here
2(2) + 1 = 5
f(2) = 5
X=4
So:one fits here
This
2(4) + 1 = 9
f(4) = 9
Now you try!
• Evaluate f(x) for f(-3), f(-1), f(5)
 2 x, ifx  1
f ( x)  
2 x  1, ifx  1
Graph:
 x  , if x  1
f ( x)  
  x  3, if x  1
1
2
3
2
•For all x’s < 1, use the top graph (to the left of 1)
•For all x’s ≥ 1, use the bottom graph (to the
•right of 1)
 12 x  32 , if x  1
f ( x)  
  x  3, if x  1
x=1 is the breaking
point of the graph.
To graph the top equation, pick
two points less than 1. Since
this is a linear function we can
just plot two points and
connect them.
To graph the bottom equation
graph, pick two points greater
than or equal to 1. Since this is
a linear function we can just
plot two points and connect
them.
Now you try to Graph:
2
2
 3 x  3 , if x  2
f ( x)  
  x  1, if x  2

Step Functions
 1, if
 2, if
f ( x)  
if
,
3

 4, if
0  x 1
1 x  2
2 x3
3 x  4
 1, if 0  x  1
 2, if 1  x  2
f ( x)  
 3, if 2  x  3
 4, if 3  x  4
You try to graph :
 1, if  4  x  3
 2, if  3  x  2
f ( x)  
3
,
if

2

x


1

 4, if  1  x  0
Graphing Piecewise functions
• Regardless of what the piecewise function
is, to graph:
– Find the ‘breaking points’ and draw a dashed vertical line.
– Plot points for the correct domain(s)
– Determine if the graph is continuous or if it has a ‘hole’.