Transcript 2.1 Functions and Their Graphs

Find the solution(s) to each equation.
1) (x-3)(x+3) = 0
2) (x-4)(x+1) = 0
x – 3 = 0 and x +3 = 0
x – 4 = 0 and x +1 = 0
x = 3 and x = -3
3) x(x2 – 1) = 0
x(x – 1)(x + 1) = 0
x = 4 and x = -1
4) x2 – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 0, x – 1 = 0, and x +1 = 0
x – 5 = 0 and x +1 = 0
x = 0, x = 1, and x = -1
x = 5 and x = -1
9.3a Rational Functions and Their Graphs
Objective – To be able to graph general rational functions.
GRAPHS OF RATIONAL FUNCTIONS
p(x)
a x m + ……
f(x) =
=
Has the following characteristics:
n
q(x)
b x + ……
vertical asymptote is at each real zero of q(x).
Example 1
Find any points of discontinuity (vertical asymptotes)
y=
3
x2 – x – 12
x2 – x – 12
(x – 4)(x + 3)
x – 4 = 0 and x + 3 = 0
+4
+4
-3
x = 4 and x = -3
-3
On White Board
Find any points of discontinuity (vertical asymptotes)
y=
1
x2 – 16
x2 – 16
(x – 4)(x + 4)
x – 4 = 0 and x + 4 = 0
+4
+4
-4
x = 4 and x = -4
-4
On White Board
Find any points of discontinuity (vertical asymptotes)
y=
x+1
x2 + 2x – 8
x2 + 2x – 8
(x + 4)(x – 2)
x + 4 = 0 and x – 2 = 0
-4
-4
+2
x = -4 and x = 2
+2
On White Board
Find any points of discontinuity (vertical asymptotes)
y=
x2 – 1
x2 + 3
x2 + 3
x2 + 3 = 0
-3
-3
x2 = -3
None
GRAPHS OF RATIONAL FUNCTIONS
p(x)
a x m + ……
f(x) =
=
Has the following characteristics:
n
q(x)
b x + ……
horizontal asymptote:
- If m < n, the line y = 0 is a horizontal asymptote.
- If m = n, the line y =
a
b
is a horizontal asymptote.
- If m > n, the graph has no horizontal asymptote.
Example 2
Find the horizontal asymptote of:
y=
– 4x + 3
2x + 1
y=
-4
/2
y= –2
**Remember that if the
exponent is the same you
use the coefficients.
On White Board
Find the horizontal asymptote of:
y=
– 2x + 6
x–1
y=
-2
/1
y= –2
**Remember that if the
exponent is the same you
use the coefficients.
On White Board
Find the horizontal asymptote of:
y=
2x + 5
x2 + 1
y= 0
On White Board
Find the horizontal asymptote of:
y=
4x5 + 5
x2 + 1
None
9.3b Rational Functions and Their Graphs
Objective – To be able to graph general rational functions.
GRAPHS OF RATIONAL FUNCTIONS
p(x)
am x m + ……
f(x) =
=
Has the following characteristics:
n
q(x)
bn x + ……
1. x-intercepts are the real zeros of p(x).
2. vertical asymptote is at each real zero of q(x).
3. horizontal asymptote:
- If m < n, the line y = 0 is a horizontal asymptote.
- If m = n, the line y =
am
bn
is a horizontal asymptote.
- If m > n, the graph has no horizontal asymptote.
Example 1
Graph
y=
4x2
x2 – 9
Where are the asymptotes?
5
4
x y
-4 9.1
VA: x = 3, x = -3
HA: y = 4/1
-5 61/4
HA: y = 4
-1
-1/
2
0 0
1
-1/
3
2
1
–5 –4 –3 –2 –1
–1
–2
2
4 9.1
5 6 1/4
–3
–4
–5
1
2
3
4
5
Example 2
Graph the function
x+1
y=
(x – 3)(x + 2)
x y
0 -1/
6
1 -1/3
-1 0
5
VA: x = 3 and x = -2
HA: y = 0
4
3
2
1
–5 –4 –3 –2 –1
–1
-3
-1/
3
–2
-4
-3/
14
–3
4
5/
6
5
6/
14
–4
–5
1
2
3
4
5
Example 3
Graph
y=
x2 – 2x +1
x–2
Where are the asymptotes?
5
4
x y
1 0
x=2
3
2
0
-1/
2
1
-1
-4/
3
–5 –4 –3 –2 –1
–1
3 4
4
5
–2
9/
2
16/
–3
–4
3
–5
1
2
3
4
5
Sec. 9.4 Multiplying and
Dividing Rational Expressions
Objective: Multiply/Divide
and Simplify Rational
Expressions
Example 1
•Simplify
2
•x
– 7x – 18
2
x – 8x -9
• First Factor top and bottom
(x – 9)(x + 2)
(x – 9)(x + 1)
Numerator
-18
-9
2
-7
Then cancel like terms and get
Bottom
-9
(x + 2)
(x + 1)
1
-9
-8
Example 2
• Simplify
5
9x
6 5x
 7
5 x 18
• See if you can cross cancel anything.
3
5
9x
6 5x
 7 2
5
x 18 2
=
3
x
Example 3
• Simplify
x  4 x  12 x  2 x  35

2
x  11x  30
x4
2
2
• First do like example 1 and factor
everything
• (x + 6)(x – 2) (x – 7)(x + 5)

(x + 6)(x + 5)
(x + 4)
• Cancel all appropriate parts
• We Get
• (x – 2)(x – 7) or x2 – 9x + 14
x+4
x+4
1st Numerator
-12
-2
6
4
1st Denominator
30
6
5
11
2nd Numerator
-35
5
-7
-2
Example 4
• Simplify
x  25
x5

2
2
x  2 x  3 x  3x  18
2
• What do we ever do when we
divide by a fraction.
• Yes we multiply by the
reciprocal
x  25 x  3x  18

2
x  2x  3
x5
2
2
• Do the same steps from before and
we get
• (x + 5)(x – 5)
(x - 6)(x + 3)

(x + 3)(x – 1)
x+5
After Cross Cancelling we end up
with
2
(x – 5)(x – 6) or x – 11x + 30
x–1
x-1
Last Example:Complex Fraction
• Simplify
b 4
2
b  2b  1
b2
b 1
2
• Just like the previous problem we write the
bottom fraction by the reciprocal and mult.
b 4
b 1

2
b  2b  1 b  2
2
• When you factor you get.
(b  2)(b  2)
b 1

2
(b  1)
b2
b2

b 1
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
An object is 24 cm from a camera lens. The object is in focus
on the film when the lens is 12 cm from the film. Find the focal length
of the lens.
1
1 1
=
+
f
di do
Use the lens equation.
1 1
1
=
+
f 12 24
Substitute.
=
2
1
+
24 24
Write equivalent fractions with the LCD.
=
3 1
=
24 8
Add and simplify.
Since 1 = 1 , the focal length of the lens is 8 cm.
f
8
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
Find the least common multiple of 2x2 – 8x + 8 and 15x2 – 60.
Step 1: Find the prime factors of each expression.
2x2 – 8x + 8 = (2)(x2 – 4x + 4) = (2)(x – 2)(x – 2)
15x2 – 60 = (15)(x2 – 4) = (3)(5)(x – 2)(x + 2)
Step 2: Write each prime factor the greatest number of times it appears
in either expression. Simplify where possible.
(2)(3)(5)(x – 2)(x – 2)(x + 2) = 30(x – 2)2(x + 2)
The least common multiple is 30(x + 2)(x – 2)2.
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
Simplify
1
4x
+
.
3x2 + 21x +30
3x + 15
1
4x
1
4x
+
=
+
3(x + 2)(x + 5) 3(x + 5)
3x2 + 21x +30
3x + 15
=
x+2
1
4x
+
• x+2
3(x + 2)(x + 5) 3(x + 5)
1
4x(x + 2)
+
3(x + 2)(x + 5)
3(x + 2)(x + 5)
1 + 4x(x + 2)
=
3(x + 2)(x + 5)
2
= 4x + 8x +1
3(x + 2)(x + 5)
=
4x2 + 8x +1
= 2
3x + 21x +30
Factor the
denominators.
Identity for
Multiplication.
Multiply.
Add.
Simplify the
numerator.
Simplify the
denominator.
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
Simplify
x2
2x
3
–
– 2x – 3
4x + 4
2x
x2 – 2x – 3
–
3 .
4x + 4
=
2x
3
–
(x – 3)(x + 1) 4(x + 1)
=
4
2x
3
• 4 –
• x–3
(x – 3)(x + 1)
4(x + 1) x – 3
Factor the
denominators.
Identity for
Multiplication.
=
4(2x) – (3)(x – 3)
4(x + 1)(x – 3)
Simplify.
=
5x + 9
4x2 – 8x – 12
Simplify.
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
Simplify
1 1
+
x y
2 1
–
y x
.
Method 1: First find the LCD of all the rational expressions.
1 1
+
x y
2 1
–
y x
=
=
1 1
+ • xy
x y
2 1
– • xy
y x
1 • xy 1 • xy
+ y
x
2 • xy
1 • xy
–
y
x
y+x
= 2x – y
The LCD is xy. Multiply the numerator
and denominator by xy.
Use the Distributive Property.
Simplify.
Adding and Subtracting Rational Expressions
LESSON 9-5
Additional Examples
(continued)
Method 2: First simplify the numerator and denominator.
1 1
+
x y
y
x
+
xy xy
=
2x
x
2 1
–
–
xy xy
y x
x+y
xy
=
2x – y
xy
x+y
2x – y
= xy ÷ xy
x+y
xy
= xy
• 2x – y
x+y
= 2x – y
Write equivalent expressions with
common denominators.
Add.
Divide the numerator fraction
by the denominator fraction.
Multiply by the reciprocal.
Solving Rational Equations
LESSON 9-6
Additional Examples
Solve
1
= 6x
. Check each solution.
x – 3 x2– 9
1
= 6x
x – 3 x 2– 9
x2 – 9 = 6x(x – 3)
Write the cross products.
x2 – 9 = 6x2 – 18x
Distributive Property
–5x2 + 18x – 9 = 0
Write in standard form.
5x2 – 18x + 9 = 0
Multiply each side by –1.
(5x – 3)(x – 3) = 0
Factor.
5x – 3 = 0 or x – 3 = 0
x=3
5
or
x=3
Zero-Product Property
Solving Rational Equations
LESSON 9-6
Additional Examples
(continued)
Check: When x = 3, both denominators in the original
equation are zero.
The original equation is undefined at x = 3.
So x = 3 is not a solution.
When 3 is substituted for x in the original equation,
5
5
both sides equal – 12 .
Solving Rational Equations
LESSON 9-6
Additional Examples
Solve
3
4 1
– = .
5x 3x 3
3
– 4 = 1.
5x 3x 3
1
15x 3 – 4 = 15x 3
5x
3x
45x – 60x = 15x
3
5x
3x
9 – 20 = 5x
–
Multiply each side by the LCD, 15x.
Distributive Property
Simplify.
11
=x
5
Since – 11 makes the original equation true, the solution is x = – 11.
5
5
Solving Rational Equations
LESSON 9-6
Additional Examples
Josefina can row 4 miles upstream in a river in the same time it
takes her to row 6 miles downstream. Her rate of rowing in still water is 2
miles per hour. Find the speed of the river current.
Relate: speed with the current = speed in still water + speed of the current,
speed against the current = speed in still water – speed of the current,
time to row 4 miles upstream = time to row 6 miles downstream
Define:
Distance (mi)
Write:
Rate (mi/h)
With current
6
2+r
Against current
4
2–r
6
4
=
(2 + r )
(2 – r )
Time (h)
6
(2 + r )
4
(2 – r )
Solving Rational Equations
LESSON 9-6
Additional Examples
(continued)
6
4
=
(2 + r )
(2 – r )
4
6
(2 + r )(2 – r ) (2 + r ) = (2 + r )(2 – r ) (2 – r )
(2 – r )(6) = (2 + r )(4)
12 – 6r = 8 + 4r
4 = 10r
0.4 = r
The speed of the river current is 0.4 mi/h.
Multiply by the LCD
(2 + r )(2 – r ).
Simplify.
Distributive Property
Solve for r.
Simplify.
k
k
Solve

2
k 1 k  2
LCD is (k+1)(k-2)
Find the LCD
Multiply everything by LCD
k
k
(k  1)(k  2)
 (k  1)(k  2)
 (k  1)(k  2)2
k 1
k 2
k  2k  k  k
2
 2k  2k  4
1
2
2
(continued)
2k  k  2k  2k  4
2
k  4
2
Combine Like Terms
Combine 2k2 and k’s to left side