1. Ch. 9.3 Rational Functions and Their Graphs

2. Rational Functions and Their Graphs
ALGEBRA 2 LESSON 9-3
For each rational function, find any points of discontinuity.
a. y = 3
x2 – x –12
The function is undefined at values of x for which x2 – x – 12 = 0.
x2 – x – 12 = 0 Set the denominator equal to zero.
(x – 4)(x + 3) = 0 Solve by factoring or using the
Quadratic Formula.
x – 4 = 0 or x + 3 = 0 Zero-Product Property
x = 4 or x = –3 Solve for x.
There are points of discontinuity at x = 4 and x = –3.
9-3

3. Rational Functions and Their Graphs
ALGEBRA 2 LESSON 9-3
(continued) 2x
3x2 + 4
b. y =
The function is undefined at values of 3x2 + 4 = 0.
3x2 + 4 = 0 Set the denominator equal to zero.
x2 = – Solve for x. 4 3
x = =
± –4 3
± 2i
Since is not a real number, there is no real value for x for
which the function y = is undefined. There is no point
of discontinuity.
± 2i 3 2x
3x2 + 4
9-3

4. Try This
For each rational function, find any points of discontinuity
x = 4 and x = -4

5. Try This No point of discontinuity
For each rational function, find any points of discontinuity
No point of discontinuity

6. Try This
For each rational function, find any points of discontinuity
x = -4 and x = 2

7. Vertical Asymptotes
The rational function has a point of discontinuity for each real zero of Q(x).
If P(x) and Q(x) have no common real zeros, then the graph of f(x) has a vertical asymptote at each real zero of Q(x).
If P(x) and Q(x) have a common real zero a, then there is a hole in the graph or a vertical asymptote at x = a

8. Rational Functions and Their Graphs
ALGEBRA 2 LESSON 9-3
Describe the vertical asymptotes and holes for the graph of each rational function.
a. y =
x – 7
(x + 1)(x + 5)
Since –1 and –5 are the zeros of the denominator and neither is a zero
of the numerator, x = –1 and x = –5 are vertical asymptotes.
b. y =
(x + 3)x
x + 3
–3 is a zero of both the numerator and the denominator.
The graph of this function is the same as the graph y = x,
except it has a hole at x = –3.
c. y =
(x – 6)(x + 9)
(x + 9)(x + 9)(x – 6)
6 is a zero of both the numerator and the denominator.
The graph of the function is the same as the graph y =
which has a vertical asymptote at x = –9, except it has a hole at x = 6. 1
(x + 9) ,
9-3

9. Vertical asymptotes: x = 1 and x = -3
Try This
Describe the vertical asymptotes and holes for the graph of each rational function.
Vertical asymptotes: x = 1 and x = -3

10. Try This
Describe the vertical asymptotes and holes for the graph of each rational function.
Hole : x = 2; V.A.: x = -3

11. Try This
Describe the vertical asymptotes and holes for the graph of each rational function.
Hole: x = -1; no V.A.

12. Horizontal Asymptotes
If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y = 0 (x-axis)
Ex.
H.A.: y = 0
Degree of 1
Degree of 2
2. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the fraction formed by the coefficients of the terms with the highest degree.
Ex.
H.A.:
Degree of 2
Degree of 2

13. Horizontal Asymptotes
If the degree of the numerator is greater than the degree of the denominator, then the graph has no horizontal asymptote.
Ex.
No horizontal asymptote

14. Rational Functions and Their Graphs
ALGEBRA 2 LESSON 9-3
–4x + 3
2x + 1
Find the horizontal asymptote of y = .
The degree of the numerator and denominator are equal
The horizontal asymptote is y = –2.
9-3

15. Try this
Find the horizontal asymptote
No H.A.

16. Try this
Find the horizontal asymptote
H.A.: y = 0

17. Try this
Find the horizontal asymptote

18. Rational Functions and Their Graphs
ALGEBRA 2 LESSON 9-3
x + 1
(x – 3)(x + 2)
Sketch the graph y =
Since 3 and –2 are the zeros of the denominator,
the vertical asymptotes are at x = 3 and x = –2.
Calculate the values of y for values of x near the asymptotes.
Plot those points and sketch the graph.
The degree of the denominator is greater
than the degree of the numerator,
so the x-axis is the horizontal asymptote.
When x > 3, y is positive. So as x increases,
the graph approaches the y-axis from above.
When x < –2, y is negative. So as x decreases,
the graph approaches the y-axis from below.
Since –1 is the zero of the numerator, the x-intercept is at –1.
9-3

19. Try This
Sketch the graph of
H.A.:
V.A.:
x-int.:
y-int.:

20. Page 497, Exercises: #2 – 18 e, 26 – 30 all
Quiz Ch next class