Merced College; Don Power
INTERMEDIATE ALGEBRA - CH 6, LECTURE
6.1
Rational Functions --
Simplifying; Domains
Basic property P / Q = PK / QK and vice versa:
building fractions up in preparation for addition/subtraction
reducing fractions in multiplication
Ex 6x / 8x reduces to 3 / 4
Divide num and denom by 2: best way for numerical fractions; doesn't work for algebraic ones
Factor and reduce: only sensible way for "rational functions" (fractions of polynomials)
To simplify, factor and reduce.
Caution: do not reduce without factoring
Domain: Set of all values of x for which the expression exists
This is an issue for rational functions, because we cannot divide by 0
(which means we cannot have a denominator of 0)
So the domain is all real numbers except values of x that make the denominator 0
Previous example: 6x / 8x is equivalent to 3 / 4, x ≠ 0
The complete task for simplifying consists of
Factor
Reduce any factors that are common to the numerator and denominator
For any cancelled factors, include a domain restriction.
[factors that don't cancel still restrict the domain, but we can still see the restriction in the answer]
Reducing when there is a sign change:
Principle: (x-a) / (a-x) = -1 (Why?)
Reduce both factors and change the sign of the result: you may
insert a negative sign
insert
remove a negative sign, or
remove a factor of –1, or
switch sides on a factor of the form (a - b) [because (b-a) has the opposite sign
Ex: 
Applications
Average Cost: #86
Finding a ratio: Like #85 (square pool vol to inscribed circular pool vol, if same depth)
Bonus: Find expression for the time traveled if distance is d and rate is r+5 or r-5.
From previous textbook:
Ex: (x+2) / (x2-x-6)
Ex: (x2-4) / (x3+8) Refresher on sum/diff of cubes
Ex: (2x4+14x3+20x2) / (2x5+4x4-50x3-100x2) Refresher: factoring by grouping
Finding function values
Ex: h(6), h(5) for h(x)=
Finding domain
Same example: state the domain, in proper set notation: {x|x¹5}
or interval notation: (-∞,5)U(5, ∞)
Graphing
Same example -- evaluate the reduced function when x=5
Compare/contrast graphs of reduced and unreduced functions: hole in graph
Graphs are the same except for the difference in domain
6.2
Multiplying and Dividing Rational Expressions
To multiply, factor and reduce.
DO NOT MULTIPLY ALGEBRAIC FACTORS: leave result factored so that you can tell (for future classes):
when is the answer undefined?
when is the answer zero? (how?)
To divide, multiply by the reciprocal of the divisor
Remember to include domain restrictions (for reduced factors) along with your answer
Examples:
Monomial factors: (6x2 / 5y3) * (11z2 / 2x2) / (33z5 / 10y8)
Common factors: [(490x2 - 640) / (49x2 - 112x +64)] * [(28x2 - 95x + 72) / (56x3 - 62x2 - 144x)]
Grouping: [(mx+ my + 2x + 2y) / (6x2 - 5xy - 4y2)] / [(2mx - 4x + my - 2y) / (3mx - 6x - 4my + 8y)]
Cubes with sign change: [(r3 - s3) / (s - r)] / [(r2 + rs + s2) / (r + s)2]
6.3 Adding and Subtracting Rational
Expressions
For two fractions, where the LCD is not obvious:
Factor both denominators into the GCF times an "uncommon factor"
Then the LCD is the GCF times both of the uncommon factors
So, multiply each fraction, top and bottom, by the uncommon factor from the other fraction
Ex: 5/28 + 11/49
Denoms factor into 7*4 and 7*7, with 7 being the GCF and the uncommon factors 4 and 7
Mult 5/28 by 7/7, and mult 11/49 by 4/4
So the problem becomes 5∙7 / (28∙7) + 11∙4 / (49∙4)
Note that the LCD is a byproduct -- we don't need it to set up the addition/subt
or: Multiply each fraction, top and bottom, by the opposite reduced denominator.
For the example, reduced denominators (after division by 7) are 4 and 7 respectively.
Finding LCD for algebraic factors: PRIME FACTORIZATION TECHNIQUE
1. Factor completely; write results with exponents, where appropriate
2. List every base once
3. For each base, copy the highest exponent
To use the LCD to set up fractions to add/subt:
1. Compare the factored denom of a fraction with the LCD
2. Identify which factors are present in the LCD but missing in the denominator
3. Multiply the fraction, top and bottom, by those missing factors.
Ex from Math 2/26/4B: A/x + B/(x+1) + C/(x+1)2 + D/(x+1)3 +(Ex+F)/(x2+2)
Ex:
(same denominator);
x / (x2-y2)
- y / (x2-y2)
(opposite signs), x / (x2-5x+6) - 3 / (3-x)
(factoring, no expo), (5x+3) / (2x2+5x+3)
– (3x+9) / (2x2+7x+6)
(with exponents), x / (x2+4x+4)
– 2 / (2+x)
(diff of cubes), 1 / (2y-3) - 18y / (8y3-27)
(polynomial + fraction), x - 5 / (3x+4)
+ 1
Word prob write an expression for the sum of the reciprocals of two consecutive integers.
-- or two consecutive even (or odd) integers
6.4 Complex Fractions
These are fractions containing fractions.
Basic Technique: A fraction is a division problem, so ...
Divide the numerator by the denominator, therefore, ...
Divide the [fraction in the] numerator by the [fraction in the] denominator
Ex: #8 monomials, simplification with exponents
#16 factoring
#24 any division problem is the same task as a complex fraction
When we have several fractions in the numerator and/or denominator, we can
Method 1: Simplify the numerator and denominator of the complex fraction separately
(usually: add/subtract);
Then finish by dividing as described above. or ...
Method 2: Multiply the numerator and denominator of the complex fraction by the LCD of all the fractions (Described in study tip on page 400)
Ex: #30 Both ways
Ex: #35 Denoms of x and x2
Ex: #46 Multiple factors in denom
Ex: #53 Neg exponents
Ex: Find the reciprocal of 1/a + 1/b (You'll need this technique for #65)
Previous textbook: Simplify the difference quotient for f(x) = 1/(x+h)
6.5
Dividing Polynomials; and
Synthetic Division
Division by monomial
Principle: (A+B)/C = A/B + A/C
Ex: 
Long division
2nd degree trinomial by binomial of form x-a:
Ex: (3x2-4x+7) / (x-2)
Check the result by multiplication
Dividing into a polynomial with missing term(s):
When do you quit? When the remainder is of a lower degree than the divisor
Note how to handle the remainder
Ex: #48 with missing terms
Consequence if remainder is 0:
Ex: (3x6+x3-14) / (x3-2) The check shows that you have factored the numerator
Ex: Factor 32x5+1 if you suspect that one of the factors is 2x+1
Information: In future classes, you learn techniques that will tell you that 2x+1 is a factor
Synthetic Division
Applies if the divisor is of the form x-k, where k is a constant.
We change the divisor to k (versus -k), so we get to add instead of subtracting
New table approach eliminates all the duplication, and even all the dividing
Vertical pattern: Add terms
Diagonal pattern: Multiply by k
Ex: #59, 66
Easy factoring by division (if one factor is known):
#70 3rd degree, followed by trinomial
#73 4th degree, followed by grouping
"Finding a pattern" in #85: Shows that for a polynomial y = f(x), if you perform synthetic division of f(x) by k, the remainder is f(k), i.e. if x = k, then y = remainder. The result can be used to build a table of values for graphing. (and it's usually much faster than using substitution).
Ex: Graph y = –3x2 + 2x +4 for all integers from –3 to 3
6.6
Solving Rational Equations
Biggies:
You must clear fractions, at least if you start with a variable in the denominator.
You must check your answers, at least to verify that the solutions don't make any denominator zero.
Clearing fractions. 3 Techniq ues
1. General procedure: multiply both sides of eqn (therefore every term) by the LCD
2. For proportions, clear fractions by cross-multiplying
Example: 
Solve 
for y:
Cross-multiply to get: x (y − 1) = y−3
Clear parentheses: xy − x = y − 3
Collect terms with y on the left: xy − y = x − 3
Factor out y: y (x − 1) = x − 3
Divide to finish: y = (x − 3 / (x − 1)
Exception: Use the general procedure if the denominators have a common factor
3. For proportions, it is sometimes faster to take the reciprocal of both sides
Example: Solve 
The
solution is x = 3/7.
[but you can't take reciprocals term by term, where either side has two or more terms]
Ex. To use this technique on 
first add the
fractions on the right: 
;
then take
reciprocals: 
6.7
Applications and
Variation
These applications are some that require solving of rational equations..
Distance-rate-time where there is a known relationship between the times
Times are equal (Example 1 in textbook), or
Total time is known (#33), or
Difference in the times is known
If d = r*t, then t = d / r, so...
Calculate expressions for both times, use the known relationship to set up the equation
The same technique works when there is a known relationship between the rates: r = d/t
Example when the difference in the rates is known: #34
Work-rate problems: Don't add the times, add the rates (e.g. jobs per hour)
Ex: Larry can do the job by himself in 3 hr, Mo in 5 hr, or Curly in 15 hr.
How long does it take if they work together? (3+5+15? Well, they are the 3 stooges.
But if they could cooperate, 1/3 + 1/5 +1/15 = 1/t
Ex: A sink can be filled in 5 min, or it can be drained in 3 min.
How long to empty the full sink if the drain is opened and the faucet is turned on?
+1/5 –1/3 = –1/t
Variation is a technique that can solve a wide variety of application problems.
Typical word pictures " y is proportional to . . . x (or x2, or x3, or sqrt(x), etc.)
y varies directly with . . .
y varies with . . .
y varies inversely with . . .
y is inversely proportional to . . .
y varies jointly with . . . two quantities
Situations represented:
Direct variation: If we double one quantity, we double the other; triple one, triple the other, etc.
Example: On a trip, if we double the amount of gasoline, we double the distance traveled
Example 2: To double the area of a pizza, we have to double the square of the radius (A=pr2)
Inverse variation: Double one qty and cut the other in half; triple one and divide the other by 3, etc.
Example: On a trip, if we double our speed, we should cut the time in half
Non-example: If we double a Celsius temp., we don't double the corresponding Fahrenheit temp.
One basic family of formulas works for all these situations
Based on the wording of the problem, we set up our own equation.
Steps:
1. Based on a variation statement, write a general equation
Typically y=kx or y=k/x (with variations on the x--see below), ...
This equation will include a variation constant k
We'll choose meaningful letters instead of the y and x
2. Calculate the value of the constant k
How? Substitute a matched set of numbers, so that k is the only remaining letter.
3. Rewrite the variation equation using the calculated value of k
Leave the other elements as variables
4. Use the variation equation to solve application problems
(all values will be known except for one letter)
Forms for:
Direct: y=kx, y=kx2, y=kx3 y=k*sqrt(x), etc. "y varies with x" "y is proportional to x" etc.
Inverse: y=k/x, y=k/x2 ("inverse square"), y=k/x3 etc.
Joint: y=k*x*z, y = k*x*z / w, etc.
modifications/combinations of these - see exercises
Tasks in textbook's exercises:
1. Set up the basic variation equation (i.e. the form above, doing just step 1 above) #1-12
2. Also find the constant of proportionality (step 2 above),
and write an equation that relates the variables (step 3 above), #21-32
3. Also apply the completed equation to a new situation,
where all but one of the variables are known #43-55, 60a, 60f
Examples:
Direct: Set-up #5, Equation #21, Applied #45
[direct variation problems can also be done by and proportion]
Inverse: Set-up #9, Equation #25, Applied #54
Joint: Set-up #12, Equation #30, 31, Applied #60a and 60f
Inverse square: Set-up #8, Equation #28, Applied #14 [joint variation, with inverse square]
McKeague:
6.1 Completing the Square
Square root principle: If a2 = b, then a = ±sqrt(b) a and b may be any algebraic expressions.
Ex: Solve (2x + 3)2 = 13
You must be able to approximate solutions with a calculator: Check previous solution with calculator
How to solve a quadratic eqn by completing the square
1. Write equation in form ax2 + bx = c
2. If a¹1, divide through by a. The equation is now in the form x2 + bx = c
3. Divide b by 2, square the result, and add to both sides.
4. The left side is now a perfect square; solve by the square root principle.
Ex: Easy if a=1, b is even: Solve x2 -8x + 3 = 0
Ex: Harder otherwise due to fractions: Solve 3x2 + 2x - 7 = 0
Special triangles - It's useful to know the ratios of the sides
45 - 45 - 90 Ratios: 1 to 1 to sqrt(2) Why? Solve for the hypotenuse in the triangle.
30 - 60 - 90 Ratios: 1 to 2 to sqrt(3) Why? Solve for the altitude in an equilateral triangle.
Alternate technique in the case a¹1 or b is not even:
1. Add b2/(4a) to both sides (or add and subtract on the same side)
2. To get a perfect square, either
a. Multiply through by 4a (or just by a if b is even)-- [avoids fractions to last step], or
b. Divide through by "a" (the coefficient of x) -- [most direct solution for x], or
c. Factor out "a" (the coefficient of x) -- best for completing the square on several expressions.
3. Solve by the square root principle.
Ex: (same as above) Solve 3x2 + 2x - 7 = 0 Take option of clearing fractions.
6.2 The Quadratic Formula
Derivation: complete the square on ax2 + bx +c = 0
Always solve by easier techniques if possible
Square root principle
Factoring
Completing the square if a=1 and b is even.
Important issue: simplifying the radical and the fraction
Ex (same as in last lesson): 3x2 + 2x - 7 = 0 gives result (-2 ± sqrt(88)) / 6
Teaching note:
Any even b will result in the need to simplify.
You also need to simplify if a contains a perfect square factor and the root is a factor of b
Ex: 9x2 - 3x = -7
Ex: 5x - 7 / (5x) = 1 Quadratic after clearing fractions: 25x2 - 5x -7 = 0
Rational equations can result in quadratics: Ex #29-36
Ex: 2/(x+2) - 4/(x-3) = 5 Sol: (3 ± sqrt329) / 10
Options for reducing (after simplifying radical):
Write as two separate fractions, and reduce each of them, or
Factor out the common factor from the numerator before reducing, or
Divide all three terms by their greatest common divisor
6.3 Additional Items Involving Solutions to Equations
Discriminant - the radicand in the quadratic equation b2-4ac
If positive -- two real solutions
If positive and a perfect square -- solutions are rational (otherwise irrational)
(you can always solve these by factoring)
If negative -- two complex solutions
If zero, one real (rational) solution. (expression is a perfect square).
Ex: 2x2-4x-1=0: discriminant is 16+4*2=24. 2 real (irrational)
2x2-5x-3=0: discriminant is 25+4*6=49. Factorable; 2 real (rational) sol.
x2+2x+7=0: discriminant is 4-4*7 = -24. Two complex solutions
9x2-12x+4=0: discriminant is 144-4*36=0. Perfect square: one real solution.
Building equations from their solutions:
Use roots (solutions) to find factors, then use factors to build an equation (not unique unless ...)
Ex: roots/solutions t=5, t= -3, t=0
corresponding factors: t-5, t+3, t.
simplest equation: (t-5)(t+3)(t)=0, or t3-2t2-15t=0
Ex: roots/solutions with fractions: x=2/3, x=-3/5 i.e. 3x = 2, 5x = -3
corresponding
factors (x-2/3) and (x+3/5) or we
can write (3x-2) and (5x+3)
equation: 15x2-x-6=0
Ex: roots/solutions t=5 (multiplicity 2), t= -3 (multiplicity 3), t=0 (multiplicity 4)
corresponding factors: (t-5)2, (t+3)3, t4
simplest equation: t4 (t-5)2(t+3)3=0
Note: Any multiple of this equation also has the
same solutions
such as 2t4 (t-5)2(t+3)3=0
Finding all solutions of a cubic equation, where one solution is known:
Use root to find factor; either factor it out or divide it out; solve the resulting reduced equation.
Ex: (like 53-60): Find all the solutions to x3+2x2+x+12 if x= -3 is one of the solutions.
Ex: t=2 is a sol. of t3=8t2-25t+26
Using graphs to find solutions (this works to find all real solutions of any equation)
Fact: the x-intercepts of the graph y=f(x) are the same as the solutions of f(x)=0
6.4 Equations Quadratic in Form
Situation: equation contains some function of x and that same expression squared.
Technique: Let a single letter represent that entire expression
Ex 1: expression is (x+3) Let y = (x+3); so y2 = (x+3)2
Ex 2: expression is (x2) Let y = (x2); so y2 = (x2)2 = x4
Ex 3: expression is sqrt(x) Let y = sqrt(x); so y2 = [sqrt(x)]2 = x
After solving for y, substitute back and solve for x
Test solutions: you can pick up extraneous solutions, especially for radicals, rationals.
Golden ratio: dividing a line segment in "extreme and mean ratio"
See extension problems like 58, 60, 62, 72, 76 (implications for graphing)
6.5 Graphing Parabolas
Goal: get a sketch: exact precision is not required, but do give specific coordinates and scale.
Techniques:
1. Point-plotting
(Example: X77 from 6.3; Solve x2-1=2x by algebra and by graphing)
Advantage: easy to understand
Disadvantages
There is no guarantee that you will find the most important points (vertex, intercepts)
Techniques
Substitution (standard)
Synthetic division by x gives a remainder of f(x), which is y
2. Key points technique
(Example: y=x2-5x+2: and show the calculation of y for x=4 using synthetic division)
a. vertex - and plot line of symmetry
Formula for x-coordinate: x=-b/(2a) Note: this is the Quadratic Formula without the ± term.
Calculate y by substitution or synthetic division
b. Does it open upward or downward? Leading Coefficient Test
c. y-intercept
General technique: Set x=0 and solve for y. Therefore, the y-intercept is the constant.
d. x-intercept(s):
May or may not exist, depending on whether the graph crosses the x-axis or not.
Technique: Set y=0 and solve.
Use factoring if possible.
If factoring is not possible: either
"Bracket" the roots: Calculate the points to the left and the right of the x-intercept. or
Calculate the x-intercepts by the quadratic formula or completing the square
e. additional points based on line of symmetry
f. If you don't have enough, get another point by point-plotting (and its "mate" by symmetry)
Additional examples: y = x2 +3x + 3 (Imaginary roots); y = 3x2 - 10x + 3 (harder vertex)
6.6 Quadratic Inequalities
Quadratic Inequalities
Example: Solve x2-x-6 > 0 (solution is developed below)
Get a zero on one side of the inequality and factor the resulting expression
For rational expressions, you may have to add or subt fractions first
Expression can only change signs when the factors of the expression are zero ("critical points")
Note: these factors could be in the numerator or the denominator
The solution will involve not only these critical points, but also the intervals in between.
Problem: We can't get the intervals directly. Ex: (x-3)(x+2)>0. Sol is not x>3, x>-2 (test x= -4)
We need a technique to count the number of positive or negative factors.
Technique: Use number line
Plot solutions to the equation (and points at which the expression is undefined)
Determine whether the