1. Rational Functions Part Two
Mathematics Precalculus with trigonometry
Fall 2016

2. Hints for successful graphing Rational Functions
0. Use your calculator to sketch the graph
1. Set x = 0. Plot the y-intercepts.
2. Find the zeros of function by setting N(x) = 0. Plot the x-intercepts.
3. Find the zeros of the denominator by setting D(x) = 0. Sketch the VA.
4. Find and sketch any HA by comparing the degree of N(x) and D(x).
5. Plot at least 1 point on either side of x-intercepts and VA.
6. Draw nice smooth curves and then say ahhhhhh…

3. Let’s examine the homework problems

4. For Review : The Analysis of f From the calculator we have
the following graph
The Analysis of f
Domain : R − { 1 }
Vertical Asymptote at x = 1
Let’s Find the Horizontal Asymptotes
Using the TI-89 Calculator
limit(y1(x), x, ∞) = 2 −
limit(y1(x), x, −∞) = 2 +
We have a HA at y = 22

5. For Review : (continued)
From the calculator we have
the following graph
To find the Zeros set the
Numerator function to 0.
Solve 2x + 1 = 0. We have
a zero at ( ½ , 0)
To find the y-intercept find f(0).
So the y-intercept is ( 1, 0)
Range R − { 2 }

6. Example 1 : The Analysis of f From the calculator we have
the following graph
The Analysis of f
Domain : R
Vertical Asymptote None
Let’s find the Horizontal Asymptotes
Using the TI-89 Calculator
limit(y1(x), x, ∞) = 0 +
limit(y1(x), x, −∞) = 0 −
You CAN cross the asymptote at numbers that are not far from x=0.

7. Example 1 : (continuing)
From the calculator we have
the following graph
Zeros − Solve 5x = 0 gives
That ( 0, 0) is the only zero
To find the y-intercept find f(0).
So the y-intercept is ( 0, 0)
Using the calculator, we have a
Min at (-.58, -1.44) and a Max
At (.58, 1.44)
Range (-1.44, 1.44)

8. Let’s talk about Asymptotes
Vertical Asymptote
→ VA when D(x) = 0
Horizontal Asymptote
→3 different possibilities for HA
The degree of N(x) is less than that of
D(x).
The HA is at y=0
The degree of N(x) is equal to that of
D(x).
The HA is at y=a/b with a and b the lead coefficients of N(x) and D(x) respectively
3. The degree of N(x) is more than that of
D(x).
There is no HA
The Calculator is a help with Horizontal Asymptotes

9. Example 2 : The Anaylsis of F From the calculator we have
the following graph
The Anaylsis of F
Domain : R because
3x2 + 1 ≠ 0
Vertical Asymptote None
Comparing the degree of numerator
And denominator polynomial, we
have that the Horizontal Asymptote
is at y = ⅔
Using the TI-89 Calculator
limit(y1(x), x, ∞) = ⅔ −
limit(y1(x), x, −∞) = ⅔ −
We have that the Horizontal
Asymptote is at y = ⅔

10. Example 2 : (continuing)
From the calculator we have
the following graph
Zeros − Solve 2x2 = 0 gives
That ( 0, 0) is the only zero
To find the y-intercept find f(0).
So the y-intercept is ( 0, 0)
Range [ 0, ⅔ )

11. Example 3 :
The Analysis of F
From the calculator we have
the following graph
Domain : R
because 3x2 + 1 ≠ 0 (ever)
Vertical Asymptote None
Zeros − Solve 2x3 = 0
gives That ( 0, 0) is the
only zero
Range : R
There are no HA! Use
your calculator to verify
However, there is a
slant asymptote!!!
The equation of the slant asymptote is equal to the quotient of the long division.

12. Facts about Slant Asymptotes
The slant asymptote is present only if the degree of N is exactly 1 more than the degree of D.
The equation of the slant asymptote is the quotient of the long division of D into N.
Slant asymptote are often called oblique asymptotes.

13. Example 3 : Continued
From the calculator we have
the following graph
The equation of the slant asymptote is equal to the quotient of the long division.
So we have y = ⅔x for the
slant asymptote.

14. Problem 1 :

15. Problem 2 :

16. Problem 3 :

17. Problem 4 :
*: what is unique
about this problem?

18. Problem 5 :
Can you work backwards???