3.4 rational functions and their graphs
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3.4 Rational Functions and Their
Graphs
SUMMARY OF HOW TO FIND ASYMPTOTESVertical Asymptotes are the values that are NOT in the domain. To find them, set the denominator = 0 and solve.
“WHAT VALUES CAN I NOT PUT IN THE DENOMINATOR????”
To determine horizontal or oblique asymptotes, compare the degrees of the numerator and denominator.
1. If the degree of the top < the bottom, horizontal asymptote along the x axis (y = 0)
2. If the degree of the top = bottom, horizontal asymptote at y = leading coefficient of top over leading coefficient of bottom
3. If the degree of the top > the bottom, oblique asymptote found by long division.
Finding AsymptotesVERTICAL ASYMPTOTES
There will be a vertical asymptote at any “illegal” x value, so anywhere that would make the denominator = 0
4352
2
2
xxxxxR
Let’s set the bottom = 0 and factor and solve to find where the vertical asymptote(s) should be.
014 xx
So there are vertical asymptotes at x = 4 and x = -1.
If the degree of the numerator is less than the degree of the denominator, (remember degree is the highest power on any x term) the x axis is a horizontal asymptote.
If the degree of the numerator is less than the degree of the denominator, the x axis is a horizontal asymptote. This is along the line y = 0.
We compare the degrees of the polynomial in the numerator and the polynomial in the denominator to tell us about horizontal asymptotes.
43
522
xxxxR
degree of bottom = 2
HORIZONTAL ASYMPTOTES
degree of top = 1
1
1 < 2
If the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at:
y = leading coefficient of top
leading coefficient of bottom
degree of bottom = 2
HORIZONTAL ASYMPTOTES
degree of top = 2
The leading coefficient is the number in front of the highest powered x term.
horizontal asymptote at:
1
2
43542
2
2
xxxxxR
12
y
43
5322
23
xxxxxxR
If the degree of the numerator is greater than the degree of the denominator, then there is not a horizontal asymptote, but an oblique one. The equation is found by doing long division and the quotient is the equation of the oblique asymptote ignoring the remainder.
degree of bottom = 2
SLANT ASYMPTOTES
degree of top = 3
532 23 xxx432 xx
remainder a 5x
Oblique asymptote at y = x + 5
The graph of looks like this:
2
1x
xf
Graph x
xQ 13
This is just the reciprocal function transformed. We can trade the terms places to make it easier to see this.
31x
vertical translation,
moved up 3
x
xf 1
x
xQ 13
The vertical asymptote remains the same because in either function, x ≠ 0
The horizontal asymptote will move up 3 like the graph does.
Strategy for Graphing a Rational Function
1. Graph your asymptotes2. Plot points to the left and right of each
asymptote to see the curve
Sketch the graph of
10532)(
xxxf
• The vertical asymptote is x = -2
• The horizontal asymptote is y = 2/5
10532)(
xxxf
10532)(
xxxf
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
Sketch the graph of:
€
g(x) =1x −1
Vertical asymptotes at??
x = 1
Horizontal asymptote at??
y = 0
Sketch the graph of:
€
f (x) =2x
Vertical asymptotes at??
x = 0
Horizontal asymptote at??
y = 0
Sketch the graph of:
€
h(x) =−4x
Vertical asymptotes at??
x = 0
Horizontal asymptote at??
y = 0
Sketch the graph of:
€
y =1x + 3
− 2
Vertical asymptotes at?? x = 1
Horizontal asymptote at?? y = 0
Hopefully you remember,y = 1/x graph and it’s asymptotes:
Vertical asymptote: x = 0Horizontal asymptote: y = 0
Or…We have the function:
€
y =1x + 3
− 2
But what if we simplified this and combined like terms:
€
y =1x + 3
−2(x + 3)x + 3
y =1 − 2x − 6x + 3
y =−2x − 5x + 3
Now looking at this:Vertical Asymptotes??
x = -3
Horizontal asymptotes??
y = -2
Sketch the graph of:
€
h(x) =x 2 + 3xx
Hole at??
x = 0€
h(x) =x(x + 3)x
Find the asymptotes of each function:
€
y =x 2 + 3x − 4
x
€
y =x 2 + 3x − 28x 3 −11x 2 + 28x
€
y =x 2
x+
3xx
−4x
€
y = x + 3 −4x
Slant Asymptote:
y = x + 3
Vertical Asymptote:
x = 0
€
y =(x + 7)(x − 4)x(x − 7)(x − 4)
Hole at x = 4
Vertical Asymptote:
x = 0 and x = 7
Horizontal Asymptote:
y = 0
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