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An introductionAn introduction
L. Waihman
A function is CONTINUOUS if you can draw the graph without lifting your pencil.
A POINT OF DISCONTINUITY occurs whenthere is a break in the graph.
Note the break in the graphwhen x=3. Why?
Look at the equation of the graph. Where is this equation undefined?
2 93
xx
We can factor the numerator and reducethe fraction to determine that the graphwill be a line; however, the undefined pointremains, so there is a point of discontinuitypoint of discontinuity here.
3 33
3
x xx
x
A rational function rational function is the quotientof at least two polynomials.
The graphs of rational functions frequentlydisplay infinite and point discontinuities.
Rational functions have vertical asymptotesand may have horizontal asymptotes as well.
Let’s look at the parent function: 1xf x
If x = 0, then the entire function is undefined.Thus, there is a vertical asymptote at x=0.
Looking at the graph, you can see that the value of thefunction , as the values of x 0 from the positiveside; and the value of the function - , as the valuesof x 0 from the negative side. These are the limits ofthe function and are written as:
0lim ( )x
f x
0
lim ( )x
f x
Domain
The domain is then limited to:
,0 0,
• To find the domain of a rational function,To find the domain of a rational function, set the denominator equal to zero.set the denominator equal to zero.
• The denominator will always be all real The denominator will always be all real numbers except those values found by numbers except those values found by solving this equation.solving this equation.
Determine the domain of these rational functions:
2
1. 2
xf x
x
2
2
2 92.
xf x
x
3
2
13.
4
xf x
x
: , 2 2,D
: ,0 0,D
: , 2 2,2 2,D
Recall that a vertical asymptote occurs whenthere is a value for which the function isundefined. This means, if there are nocommon factors, anywhere the denominator equals zero.
Since a rational function is a quotient of two,polynomials, there will always be at least one value for which the entire function isundefined.
Remember that asymptotes are lines. When you label a vertical asymptote, you must write the equation of the vertical line.
Just make x equal everything it couldn’t in the domain.
State the vertical asymptotes:
2
1. 2
xf x
x
3
2
12.
4
xf x
x
V.A. : x 2
V.A. : x 2
But why isn’t this the same as our discontinuities?
Because discontinuities occur EVERYWHERE the function is undefined
VA’s exist ONLY where a factor of the numerator and denominator DO NOT cancel each other2 9
3xx
3 33
3
x xx
x
1
3x
Horizontal Asymptotes
Given: is a polynomial of degree n , is a polynomial of degree m , and , 3 possibleconditions determinea horizontal asymptote:
f x g x f xg x
•BOBO: If n<m, then is a horizontal asymptote.•If the exponent is Bigger On Bottom, the HA is y = 0
•BOTN: If n>m, then there is NO horizontal asymptote.•If the exponent is Bigger On Top, there is NO HA
• EATS DC: If n=m, then is a horizontal asymptote, where c is the quotient of the leading coefficients.•Exponents Are The Same – Divide the Coefficients
0y
y c
Find the horizontal asymptote:
x
. f xx
2 11
2
x. f x
x
3
2
12
x
. f xx x2
23
20
H.A. : y 2
H.A. : none
H.A. : y 0
One final type of asymptote
The Slant Asymptote
ONLY occurs when the NUMERATOR is one degree higher than the DENOMINATOR
Graph: 22 3
( )1
x xf x
x
Notice that in this function, the degree of the numeratoris larger than the denominator. Thus n>m and there is nohorizontal asymptote. However, if n is one more than m,the rational function will have a slant asymptote.
To find the slant asymptote, divide the numerator by the denominator:
2
2
2 5
1 2 3
2 2
5
5 5
5
x
x x x
x x
x
x
The result is . Notice that as the values of x increase, the fractional part decreases (goes to 0), so thefunction approaches the line . Thus the lineis a slant asymptote.
512 5 xx
2 5x 2 5y x
To graph a function, then
1st, find the vertical asymptote(s). (Where is the function undefined? Set denom. = 0)
2nd , find the x-intercept (s). (Set the numerator = 0 and solve)
3rd , find the y-intercept(s) . (Set x = 0 and solve)
4th , find the horizontal asymptote. (BOBO, BOTN, or EATS DC)
6th , sketch the graph.
5th , find the slant asymptote. (Is the numerator ONE degree higher than the denominator? Divide.)
Suppose that you were asked to graph:
3 51
xf xx
1st, determine where the graph is undefined. (Set the denominator to zero and solve for the variable.)
2nd , find the x-intercept by setting the numerator = to 0and solving for the variable.
1x
So, the graph crosses the x-axis at
5,0
3
1 0
1
x
x
There is a vertical asymptotevertical asymptote here.
3 5 0
3 5
5
3
x
x
x
Draw a dotted line at:
Draw a dotted line at: 3y
0, 5
3rd , find the y-intercept by letting x=0 and solving for y.
3(0) 5
0 15
1
y
y
4th , find the horizontal asymptote. (Recall the test; , so .) Thus, .
n m
y c 31
c
3y
So, the graph crosses the y-axis at
The horizontal asymptote is:
Now, put all the information together and sketchthe graph:
Graph:
4 35
xf xx
1st, find the vertical asymptote.
2nd , find the x-intercept .
3rd , find the y-intercept.
5x
3,0
4
30,
5
4th , find the horizontal asymptote. 4y
5th , sketch the graph.
Graph: 2
2
2( )
6
x xf x
x x
1st, factor the entire equation:
1 2
3 2
x x
x x
2nd , find the x-intercepts:
3rd , find the y-intercept:
3
2
x
x
1,0
2,0
10,
3
4th , find the horizontal asymptote: 1y
5th , sketch the graph.
Then find the vertical asymptotes:
Graph: 22 3
( )1
x xf x
x
Notice that in this function, the degree of the numeratoris larger than the denominator. Thus n>m and there is nohorizontal asymptote. However, if n is one more than m,the rational function will have a slant asymptote.
To find the slant asymptote, divide the numerator by the denominator:
2
2
2 5
1 2 3
2 2
5
5 5
5
x
x x x
x x
x
x
The result is . Notice that as the values of x increase, the fractional part decreases (goes to 0), so thefunction approaches the line . Thus the lineis a slant asymptote.
512 5 xx
2 5x 2 5y x
Graph:
22 3
1
x x
x
1st, find the vertical asymptote.
2nd , find the x-intercepts: and
3rd , find the y-intercept:
4th , find the horizontal asymptote. none
0,0
0,0
3,0
2
1x
2 3
1
x x
x
5th , find the slant asymptote: 2 5y x
6th , sketch the graph.
