Math 152: Blitzer quick guide and summary

• 1.1 : Algebraic expressions, Real numbers, and interval notation

Algebraic Expressions

Ex: 2x, 12 + x,√x− 1

Ex: Evaluate when t = 3:(a) 2t− 3(b) 2(1− t) + 7

Order of operations!

P - start inside, work your way outEMD - left to right!AS - left to right!

Ex: Evaluate when x = 2, y = 5.

(1)2x− 7y + 3

2y − x(2) x− 3(y + 1)2

Translating to Algebraic Expressions and equations.

Ex: Write the associated algebraic expression

(1) The product of 2 and a number.(2) The sum of a number and 3.(3) One less than a number.(4) The product of 4 and a number, subtracted from 3.(5) Two less than three times a number is 13.(6) The quotient of 3 and a number is 2.

Sets of numbers.

(1) The set of Natural numbers and the set of Whole numbers:(2) The set of Integers:(3) The set of Rational numbers:(4) The set of Irrational numbers:(5) The set of Real numbers:

“Set builder notation” and “interval notation” for SETS.

Ex: Convert between notations

1

(1) {x| x > 5}(2) (−∞, 0)(3) {x| 5 < x < 8}(4) {x| x ≥ −1}

• 1.2 : Operations with numbers and simplifying expressions.

The number line and absolute value.

Ex: Compute

(1) | − 5| (2) | − 5|

(3) −|5| (4) −(−(−| − 1|))

Adding an subtracting with the number line.

Ex: Compute using number line.

(1) 3 + 2(2) 5− 2(3) 3− 7(4) −2− 5

Add/Sub fractions and decimals.

Ex: Compute

(1)1

7− 9

7(2)

2

3+

1

5(3) 5

1

2− 2

(4) 2.34 + 17.5 (5)1

5+ .7 + 3

Mult/div real numbers, powers, and why is (−7)(−3) = 21?

Ex:

(1) (−1)(3)(−5) (2)0

2(3)

5

0(4)

(2

5

)3

(5) −32

(6) (−3)2 (7) (−2)5 (8) (−1)831

Mult/Div fractions

Ex: Simplify

(1)1

2· 3

4(2)

1

2÷ 3

4Why “flip” and multiply?? (3)

1234

2

Properties of Real numbers

CommutativeAssociativeDistributive

Combining like terms

Why is it true that 4x+ 5x = 9x ??

4x+ 5x= 4x+ 5x= x(4 + 5)= x(9)= 9x !!

Ex: Simplify

(1) 4x− 7a+ 5ab+ 6− 5x+ a− 1 (2) 3x2 − 5yx+ 7x2y + 2x− 5xy + yx2

Simplifying expressions

Ex: Simplify

(1) 2(7− x) + 5x (1) 6− 2(s+ 3) + 2s

• 1.3 : Graphing

Vocab: x-axis, y-axis, origin, quadrants

Plotting points

Ex: (2, 3), (5, 0), (−1, 2), (0,−2)

Graphing a few things with points

Ex: y = 4− x2, y = 2x+ 1, y = |x|Note: The graph is visual representation of all the points that are solutions for an equation.

• 1.4 : Solving Linear Equations

Ex: Linear equation in one variable has the form ax+ b = 0.

”Addition and multiplication properties of equality.”

3

Ex: Solve

(1) 2x+ 3 = 17 (2) 2x− 7 + x = 3x+ 1 + 2x

(3) 4(2x+ 1)− 29 = 3(2x− 5) (4)2x+ 5

5+x− 7

2=

3x+ 1

2

(5)x+ 1

3= 5− x+ 2

7(6) x+ 3 = x+ 4− 1 and x = x+ 7

(7) Suppose a hippopotamus is floating down a river at 3 miles-per-hour, then the distancefloated by the hippo after t hours is D = 3t. If the hippos favorite grass patch is exactly 18miles downstream, how long before the hippo gets there? (show work)

• 1.5 : Application questions and manipulating formulas

(1) I want to mark off an area for a garden which is three times longer than it is wide. If Ihave a spool of fencing that is 100 feet long, what dimensions should my garden be?

(2) Solve the equation 2l + 2w = p for “ l ”. This equation gives the relation between thelength, width, and perimeter of a rectangle.

(3) Solve the equation V =4

3πr3 for “ r ”. This equation gives the relation between the

radius and the volume of a sphere.

(4) Circle: formulas for area and circumference.

Formulas for volume of sphere, cube, cylinder.

• 1.6 : Exponents

(1) xa · xb = ?? (2)pf

pg= ?? (3) (bm)n = ??

(4) (rs)l = ?? (5)

(p

q

)r

= ?? (6) a−n = ??

(7)1

a−n= ?? (8)

x

0= ?? (9)

0

5= ??

4

(10) 00 = ?? (11) a0 = ?? (12)

(25x2y4

−5x6y−8

)2

SECOND WEEK:

• 1.7 : Scientific notation

Ex: 2.7× 103 , 3.0019× 10−7

To take a number out of scientific notation you

(i) Move the decimal point one place for each exponent of ten. Move to the right for positiveexponents and to the left for negative exponents.

Ex: Write as a decimal

(1) 2.7× 103 (2) −3.0019× 10−7

(3) −2.001× 102 (4) 5× 10−4

To put a number in scientific notation you

(i) Put a decimal to the right of the first non-zero digit.

(ii) Count how many places you moved the decimal point (this is the exponent on the ten).

(iii) Drop all zeros after the last non-zero digit and before the first non-zero digit.

Ex: Put into scientific notation

(1) 20, 054, 000 (2) .00705 (3) .0001

Ex: Compute

(1)(2.7× 103

) (5× 10−4

)(2)

.0006

.000002

• 2.1 : Intro to functions

Vocab: relation (ordered pair), domain, codomain, function (each element in domain goes toexactly one element in the codomain), range.

Domain: The collection of all objects for which a function is defined i.e. the collection ofinputs of a function.

Codomain: The collection of objects into which all of the outputs of a function fall.

Function: A rule that assigns each object in the domain to exactly one object in thecodomain.

Range: The collection of all outputs of a function.

Ex: (1) “The favorite color function,” f : {People in class} → {colors}

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Why does this example require that we each have only one favorite color?

(2) y = 2x+ 1

Also, sometimes written as f(x) = 2x+ 1, where f : {x− values} → {y − values}

(3) “NBA champ function,” f : {years since 1947} → {NBA teams}{2015, 2014, ..., 1975, ...} → {GSW, SAS,...}Why would we not be able to make a function in the other direction?

NOTATION:

From Ex: (1) f(Lenny) = Red , f(??) = Purple

From Ex: (2) f(0) = 1 , f(??) = 3

From Ex: (3) f(2015) = GSW , f(??) = SAS

We say, “f of 0 is 1,” i.e. the function evaluated at 0 is eual to 1.

Note: This is NOT multiplication, we are NOT multiplying “f” by 0.

• 2.2 : Graphs of functions

Graph using tables: f(x) = 2x+ 1 , g(x) = x2

“Vertical line test”

NOT functions: x2 + y2 = 1 , y2 = x

Identify domain and range from graph:

f(x) = 2x+ 1, f(x) = x2, f(x) = lnx, f(x) = ex, f(x) =√x

• 2.3 : The “Algebra of functions”

(1) (f + g)(x) = f(x) + g(x)

(2) (f − g)(x) = f(x)− g(x)

(3) (fg)(x) = f(x) · g(x)

(4)f

g(x) =

f(x)

g(x)

Ex: Perform each operation for f(x) = 2x and g(x) = x− 1

Ex: Let p(x) = x2 − 3 and q(x) = 4x+ 5, find

6

(1) (p+ q)(3) (2)q

p(0) (3) (pq)(1)

Two cool/important items:

(1) Whenever you divide functions you may have values which are not allowed in the domain,

e.g.q

p(x) = ?? and

f

g(x) = ?? , what values are not in the domain of these functions?

(2) Recall adding fractions! Ex: (a)2

3+

1

5

Now (b)5

x− 7

x− 8(c)

3

x(x+ 1)+

1

x+ 1(d)

2

x3+

1

x(x+ 1)

• 2.4 : Linear functions and slope

Graphing using intercepts

(1) To find the x-intercept, set y = 0 and solve for x.

(2) To find the y-intercept, set x = 0 and solve for y.

(3) Plot the two points and draw your line through them!

Ex: Graph

(1) 4x− 3y = 6 (2) y = 3x− 2

The slope of a line through two points (x1, y1) and (x2, y2) is

m =Rise

Run=

change in y

change in x=y2 − y1x2 − x1

Note: The order you subtract matters!

Ex: Find the slope of the line passing through the two points.

(1) (−3,−4) and (−1, 6) (2) (−1, 3) and (−4, 5)

Equation of a line in slope intercept form

Ex: Graph y = 2x+ 1 using intercepts, notice slope and y-intercept

The slope intercept form of a line is y = mx + b, where “m” is the slope and “b” is they-intercept.

Where did this come from anyways?! I’m glad you asked...

Start with m =y2 − y1x2 − x1

.

Now instead of choosing (x1, y1) and (x2, y2) as arbitrary points, choose (0, b) and (x, y).Notice (0, b) is the y-intercept.

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Then m =y − bx− 0

. Solving for y we get our wonderful formula y = mx+ b.

Ex: Put into slope intercept form (we want to do this because it makes graphing easy)

(1) 4x+ 2y = 10 (2) 3x− y = 2 (3) 2y − x = 1

Graphing using slope and y-intercept

How to...

(1) Plot the y-intercept

(2) Start at the y-intercept and use the slope for find a second point.

(3) Draw the line through these two points.

Ex: Graph

(1) y = 2x+ 1 (2) y = −x y =3

2x− 2

Graphing horizontal and vertical lines

How to...

(1) Write down two points with “y = 2” or “x = −1”

(2) Plot those points

(3) Draw your line

Ex: Graph

(1) y = 2 (2) x = −1 (3) y = 0 (4) x = 0

Slope as rate of change

Think of the y-axis as elevation gain and the x-axis as distance traveled. Now, you areclimbing up a mountain! The steeper the slope, the faster you change elevation.

Ex: Graph and discuss y = x , y = 2x , y = 10x , y =1

2x , y =

1

10x

The average rate of change of a function

Suppose I throw a basketball into the air. Using the same x and y axis as directly above, wesee that the graph of the basketball’s path is not a straight line and it is difficult to tell theslope at a given point. We can however figure out the slope of the line connecting any twopoints on the graph. This slope is the “average slope” of the graph between these two points.This is the “average rate of change” of a function.

Ex: Find the average rate of change of the function y = −(x−4)2+16 between x = 1 and x = 4

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• 2.5 : The point-slope form of the equation of a line

Point-slope form of a line is y− y1 = m(x− x1), where m is the slope and the line passesthrough the point (x1, y1).

Ex: Graph y = 2x+ 1 using slope and y-intercept.

Use the above equation choosing a point on the line and using the slope (m = 2) and seewhat you get (try a couple points from the line).

When asked to find the equation of a line use the point-slope form of a line.

(1) Find the equation of the line with slope 5 passing through the point (2, 3).

(2) Find the equation of the line passing through (0, 1) and (3, 7).

Final note: This “formula” also came right from the definition of slope!

Again we start with m =y2 − y1x2 − x1

.

Instead of choosing (x1, y1) and (x2, y2) as arbitrary points, choose (x1, y1) and (x, y). Now(x1, y1) is the point we are given. Instead of choosing a second point (x2, y2) and computingthe slope, we let the second point “vary” as (x, y), our independent variable x and ourdependent variable y.

Then m =y − y1x− x1

. Re-arranging, we get our wonderful formula y − y1 = m(x− x1).

Parallel and perpendicular lines

Slope of parallel lines. Line l1 has slope m, then any parallel line l2 also has slope m.

Ex: Find another line parallel to y = 2x+ 1.

Slope of perpendicular lines. Line l1 has slope m, any perpendicular line l2 has slope − 1

mi.e. it is the “negative reciprocal”. Another way to check if two lines are perpendicular is if(m1)(m2) = −1 . Notice that (m)

(− 1

m

)= −1 .

Ex: Find another line perpendicular to y = 2x+ 1.

• 4.1 : Solving linear inequalities

A linear inequality behaves almost exactly the same as a linear equation, except there aretwo important differences.

(1) If you multiply or divide by a negative number, you have to change the direction of theinequality sign.

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Check it out: 6 < 10 is a perfectly fine and true statement, however, if we divide both sidesby −2 we get: −3 < −5. This statement is false! In fact, −5 < −3.

(2) The solution is not a single number, but a set of numbers. If we solve an equation weend up with something that looks like x = 2, however, if we solve an inequality we end upwith something that looks like x < 2.

Give me a few numbers that make this inequality true.

This may seem like a subtle point, but x < 2 is not the answer! This is a set that we candescribe using interval notation i.e. (−∞, 2) or set builder notation i.e. {x|x < 2}.Ex: Solve

(1) 3x− 5 > −17 (2) −2x− 4 > x+ 5

Inequalities with strange solution sets

Ex: Solve

(1) x > x+ 1 (2) x < x+ 1

In the first case we write “ ∅ ”, in the second case we write “ (−∞,∞) ”.

• 4.2 : Compound Inequalities

Ex: x > 2 or x < 0 , −2 < x < 1

Intersection of Sets

The “intersection” of two sets A and B is written A ∩ B. Something is in the intersectionof two sets if it is in both sets. Think of an intersection of two streets, the intersection is onboth streets!

In set builder notation: A ∩B = {x|x ∈ A and x ∈ B}Ex: Given A = {1, 5, 9,−3, 2, 7} and B = {5, 4, 2, 7, 3}, find Find A ∩B.

Compound inequalities involving “and” i.e. “intersection”

A number is a solution to the compound inequality

x ≤ 6 and x ≥ 0

if it is a solution to both i.e. the number 1 is a solution to the compound inequality since1 ≤ 6 and 1 ≥ 0 are both true statements.

Note: You can always rewrite a compound inequality involving and in a shorter and nicerformat!

How to solve compound inequalities involving “and” or “intersection of sets”:

(1) Solve each inequality separately.

(2) Graph each inequality on separate number lines, one below the other.

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(3) See where the graphs overlap, the overlap is your solution set.

Ex: Solve

(1) x− 3 < 5 and 2x+ 4 < 14 (2) 2x− 7 > 3 and 5x− 4 < 6

Union of Sets

The “union” of two sets A and B is written A ∪B. Something is in the union of two sets ifit is in either sets. Think of a union of two families (through a wedding or such), a personis in the new bigger family if they are in either family!

In set builder notation: A ∪B = {x|x ∈ A or x ∈ B}Ex: Given A = {1, 5, 9,−3, 2, 7} and B = {5, 4, 2, 7, 3}, find Find A ∩B.

Compound inequalities involving “or” i.e. “union”

A number is a solution to the compound inequality

x ≥ 6 or x ≤ 0

if it is a solution to either i.e. the number 1 is a solution to the compound inequality since1 ≤ 0, even though 1 is not a solution to the other inequality x ≥ 6.

How to solve compound inequalities involving “or” i.e. “union of sets”:

(1) Solve each inequality separately.

(2) Graph each inequality on separate number lines, one below the other.

(3) See what part of the number line is covered by either graph, this is your solution set.

Ex: Solve

(1) 2x− 7 > 3 or 5x− 4 < 6 (2) x− 3 < 5 or 2x+ 4 < 14

• 4.3 : Equations and inequalities involving absolute value.

Ex: Solve

(a) |x| = 2 (b) |x| < 2 (c) |x| ≥ 2

Absolution Value Equalities

Ex: (a) |x| = 2 , here we have two solutions (b) |x| = −2 , here we get ???

Ex: Solve

(1) |2x−3| = 11 (2) 5|1−4x|−15 = 0 (3) |3x−1| = |x+5|

Absolution Value Inequalities type 1

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Ex: (a) |x| < 2 (b) |x| < −2 , here we get ???

Ex: Solve

(1) |x− 4| < 3 (2) −2|3x+ 5|+ 7 ≥ −13

Absolution Value Inequalities type 2

Ex: (a) |x| > 2 (b) |x| > −2 , here we get ???

Ex: Solve

(1) |2x+ 3| ≥ 5

Beware these weird solution sets:

(1) |x+ 1| < −2 (2) |x+ 1| > −2

• 5.1 : Intro to polynomials

Vocab: term, coefficient, degree of a polynomial, leading term, leading coefficient, monomial,binomial, trinomial, polynomial

End behavior of polynomial functions

Ex: Graph f(x) = x2, f(x) = x3, f(x) = −x2, f(x) = −x3

The leading coefficient criteria: As x increases or decreases without bound, the graph ofa polynomial function eventually shoots upward or plummets downward.

(a) For polynomials whose degree is even:

*If the leading coefficient is positive then (i) the graph will eventually shoot upward as xincreases and (ii) the graph will eventually shoot upward as x decreases.

*If the leading coefficient is negative then (i) the graph will eventually plummet downwardas x increases and (ii) the graph will eventually plummet downward as x decreases.

(b) For polynomials whose degree is odd:

*If the leading coefficient is positive then (i) the graph will eventually shoot upward as xincreases and (ii) the graph will eventually plummet downward as x decreases.

*If the leading coefficient is negative then (i) the graph will eventually plummet downwardas x increases and (ii) the graph will eventually shoot upward as x decreases.

Adding/subtracting polynomials

(1) (5x3y − 4x2y − 7y) + (2x3y + 6x2y − 4y − 5)

(2) (6x2 + 5x− 3)− (4x2 − 2x− 7)

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• 5.2 : Multiplication of polynomials

Ex: Multiply

(1) (5x3y4)(−6x7y8) (2) 4x3(6x5 − 2x2 + 3)

(3) (7x+ 2)(4x+ 5) (4) (x2 + 4x+ 5)(3x+ 7)

“Special Products”

(1) (a+ b)2 (2) (a− b)2 (3) (a+ b)(a− b)(4) (a+ b)(a2 − ab+ b2) (5) (a− b)(a2 + ab+ b2)

Ex: Multiply polynomial functions

(1) Given f(x) = x− 5 and g(x) = x− 2 , find (fg)(x)

Ex: Given f(x) = x2 − 7x+ 3 find

(1) f(a+ 4) (2) f(a+ h)− f(a)

• 5.3 : Greatest common factor and factor by grouping

The goal of factoring is to turn a sum of terms into a product of factors. Since we cancancel factors in a fraction, this will be super useful!

**The first thing to look for ALWAYS when factoring, is the greatest common factor.

Ex: Factor

(1) 21x2 + 28x (2) 9x2 + 15x3 (3) 12x5y4 − 4x4y3 + 2x3y2

Ex: Factor “by grouping”

(1) 2(x− 7) + 9a(x− 7) (2) 5y(a− b)− (a− b)(3) x3 − 5x2 + 3x− 15 (4) 3x2 + 12x− 2xy − 8y

• 5.4 : Factor trinomials

Ex: x2 + 2x+ 1 = (x+ 1)(x+ 1)

How to... factor trinomial with leading coefficient one

(1) Factor out GCF

(2) Determine sign of factors

(3) List factors of last term, find factors that add up to middle term

13

Ex: Factor

(1) x2 + 5x+ 6 (2) x2 − 14x+ 24 (3) y2 + 7y − 60

(4) x2−4xy−21y2 (5) 8x3−40x2−48x (6) 6y4 +12y2 +6

(7) y4−12y3+35y2 (8)−x2−4x+45 (9) x3y−2x2y2−3xy3

How to... Factor trinomials by grouping (leading coefficient not one)

(1) Multiply leading coefficient and constant.

(2) Find factors of “ac” whose sum is “b”.

(3) Re-write middle term and factor by grouping!

Ex: Factor

(1) 15x2 − 7x− 2 (2) 12x2 − 5x− 2 (3) 6x2 + 19x− 7

(4) 2x2 − 7xy + 3y2 (5) 2x2 − x− 6 (6) 8x2 − 22x+ 5

• 5.5 : Factoring special forms

Recall: a2 − b2 = (a+ b)(a− b)Ex: Factor

(1) x2 − 9 (2) 9x2 − 100 (3) 36y6 − 49x4

What about...? x2 − 5, x7 − 25, 1− x6y4, 3y − 3x6y5

Recall: a2 + 2ab+ b2 = (a+ b)2 and a2 − 2ab+ b2 = (a− b)2

Ex: Factor

(1) x2 + 14x+ 49 (2) 4x2 + 12xy + 9y2 (3) 9y4 − 12y2 + 4

What about...? x2 − 8x+ 16− y2, a2 − b2 + 10b− 25 (factor twice!)

Recall: a3 + b3 = (a+ b)(a2 − ab+ b2) and a3 − b3 = (a− b)(a2 + ab+ b2)

Ex: Factor

(1) x3 + 125 (2) x6 − 64y3 (2) 27− y3

• 5.6 : A general factoring strategy

(1) Factor out GCF from polynomial.

(2) Is the polynomial only two terms? Try a special form.

(3) Is the polynomial three terms?

(a) Try a special form, then trial/error using the list of factors.

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(b) If the coefficient on the “ x2 ” term is not 1, factor by grouping i.e. the “ ac ” method

(4) Is the polynomial four terms? Factor by grouping.

(5) Make sure it is “factored completely.”

(6) Check your answer by multiplying it out.

Ex: Factor

(1) 5x3 − 20x (2) 7x4 − 7 (3) x3 + 2x2 − 9x− 18

(4) −6x2 + 6x+ 12 (5) 8x5 − 2x3 (6) y2 + 8y − 16 prime

(7) 12x3y − 12xy2 (8) x3 − xy2 + x2y − y3 (9) 16x4y − y5

• 5.7 : Polynomial equations and applications

Ex: 2x2 + x− 1 = 0

In order to solve polynomial equations, we use the zero product principle:

if ab = 0 then a = 0 or b = 0

How to... solve polynomial equations by factoring

(1) Re-write the equation in standard form ax2 + bx+ c = 0.

(2) Factor completely

(3) Apply “sero product principle” i.e. set each factor containing a variable to zero

(4) Solve the equations in step (3)

(5) Check your solutions in original equation

Ex: Solve

(1) (3x−2)(x+2) = 0 (2) 2x2+7x−4 = 0 (3) 2x2−5x = 12

(4) x2 = 6x+ 4 (5) 5x2 = 20x (6) (x− 7)(x+ 5) = −20

(7) x3 + x2 = 4x+ 4

Modeling motion

You Throw a ball straight up on the roof of a building, the ball misses the roof and falls allthe way down to the ground.

If the function h(t) = −16t2 + 32t + 384 describes the hight of the ball after t second, howlong will it take for the ball to hit the ground?

Landscape question

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A rectangular garden measures 80ft by 60ft. You want to add a large path along both shortersides and one of the longer sides of the garden. By adding the path, you want to double thearea of the garden. How wide should the path be? (call width of path x and draw diagram)

• 6.1 : Rational Expressions and Functions intro; multiplying; dividing

Ex:13x

x− 1,

x+ 2

x2 + 2x− 1,

x+ 1

1− 2y

Domain of rational functions

Ex: Find domain

(1) f(x) =4

x− 2(2) f(x) =

2x+ 1

x2 − x− 2(3) f(x) =

2

x2 + 1

Graphs of rational functions

Ex: Graph

(1) y =1

x(2) y =

1

x− 2(3) y = −1

x

Simplifying rational expressions

How to...

(1) Factor numerator and denominator completely.

(2) Divide any common factors

Ex: Simplify

(1)5x+ 35

20x(2)

x3 + x2

x+ 1(3)

x2 + 6x5

x2 − 25

(4)x2 + 4x+ 3

x+ 1(5)

x2 − 7x− 18

2x2 + 3x− 2(6)

3x2 + 12xy + 15y2

6x3 − 6xy2

Notice: Graph f(x) =x2 + 4x+ 3

x+ 1and g(x) = x+ 3 on different graphs.

Watch out! ... Simplify(2x+ 5)(2x− 5)

3(5− 2x)

(5− 2x)= 5 + (−2x)= −2x+ 5= (−1)(2x− 5)

Multiply rational expressions

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Ex:x2

y + 3· x+ 5

y − 7=

x2(x+ 5)

(y + 3)(y − 7)

How to...

(1) Factor all numerators and denominators completely

(2) Divide all common factors

(3) Multiply remaining factors

Ex: Multiply

(1)7

x+ 3· x− 2

5(2)

x+ 3

x− 4· x

2 − 2x− 8

x2 − 9(3)

4x+ 8

6x− 3x2· 3x2 − 4x− 4

9x2 − 4

Dividing rational expressions

Recall:P

Q÷ R

S=P

Q· SR

=PS

QRWhy...?

Recall thatx

7÷ 6

yis just another way to write....

x

76

y

Ex: Divide

(1)x

7÷ 6

y(2) (4x2 + 3x− 10)÷ 2x+ 5

14

(3)x2 + 3x− 10

2x÷ x2 − 5x+ 6

x2 − 3x

• 6.2 : Add / subtract rational equations

Same denominators

Recall:2

9+

5

9=

2 + 5

9=

7

9

Ex: Add / Subtract

(1)2x− 1

3+x+ 4

3(2)

x2

x2 − 9+

9− 6x

x2 − 9(3)

2x+ 3

x+ 1− x+ 2

x+ 1

(4)x2 + 2x− 2

x2 + 3x− 10+

5x+ 12

x2 + 3x− 10(5)

x2

x− 5+

4x+ 5

5− x(6)

3y3 − 5x3

x2 − y2−4y3 − 6x3

x2 − y2

Different denominators? ... Find the Least Common Denominator (LCD)

(1) Factor each denominator completely

17

(2) List all the factors in 1st denominator

(3) Put any other factors from the 2st denominator on the list

(4) Multiply all factors on the list, this is the LCD.

Ex: Find LCD

(1)3

10x2,

7

15x(2)

9

7x2 + 28x,

11

x2 + 8x+ 16

Ex: Add / subtract

(1)3

10x2+

7

15x(2)

9

7x2 + 28x+

11

x2 + 8x+ 16

(3)x

x+ 3+x− 1

x+ 3(4)

x− 1

x2 + x− 6− x− 2

x2 + 4x+ 3

• 6.6 : Rational Equations

Ex: Solve

(1)x+ 4

2x+x+ 20

3x= 3 multiply by the LCD to clear fractions

Notice: “x” cannot be zero, but the answer is 4 6= 0 so we are okay!

(2)x+ 1

x+ 10=x− 2

x+ 4(3)

x

x− 3=

3

x− 3+ 9

(4)x

3+

9

x= 4 (5)

2x

x− 3+

6

x+ 3=−28

x2 − 9

• 6.7 : Formulas and applications

Solve for a variable

(1) Solve s− c

1− rfor “ r ” (2) Solve

1

p+

1

q=

1

ffor “ p ”

Questions involving average cost

Ex: A company has to pay $5000 rent each month, and it costs $100 to produce each what-chama-call-it.

(1) Write an equation for the cost of producing x what-chama-call-its.

(2) Write an equation for the average cost of producing x what-chama-call-its.

(3) How many what-chama-call-its must be produced each month to keep the average costdown to $200 per unit.

Questions using motion

18

Ex: You commute to work 40miles and return home the same route. Your average ratedriving home is 30 mph faster than your average rate driving to work. If the round trip takes2 hours, what is your average rate driving to work?

Distance = (rate)(time) , also, Time =distance

rateSince we want to find rate, this will be our variable we have to solve for.

Table: Distance / Rate / Time=D/RTo work: 40 x 40/xFrom work: 40 x+ 30 40/(x+ 30)

Then, 40/x+ 40/(x+ 30) = 2 since the round trip takes 2 hours.

Questions using work

You can design a webpage in 15 hours. A friend can design a webpage in 10 hours. How longwould it take to design a webpage working together?

Table: Rate / Time / Work doneYou: 1/15 x x/15Friend: 1/10 x x/10

Then x/15 + x/10 = 1 where 1 is the whole webpage you are trying to make.

• 6.3 : Simplifying complex rational expressions

How to...

(1) Find the LCD of smaller fractions

(2) Multiply numerator and denominator of larger fraction by LCD

(3) After simplifying you should have only one fraction

(4) Factor and simplify more if possible

Ex: Simplify

(1)

1

x+

y

x21

y+

x

y2

(2)

x

y− 1

x2

y2− 1

(3)

1

x+ h− 1

xh

• 6.4 : Division of polynomials

Dividing a monomial by a monomial

How to...

(1) Divide coefficients

(2) Cancel factors

Ex: Divide

19

(1)25x12

5x4(2)−12x8

4x2(3)

2x3

8x3(4)

15x5y4

3x2y

Dividing a polynomial by a monomial

How to... “Divide each term in the numerator by the denominator”

Ex: Divide

(1) (15x3 − 5x2 + x)÷ (5x) (2)10x8 + 15x6

5x3(3)

16x5 − 9x4 + 8x3

2x3

Dividing a polynomial by a binomial

How to... “This is exactly the same as high school long division”

Ex: Divide

(1) (x2 − 14x+ 24)÷ (x− 2) (2) (4− 5x− x2 + 6x3)÷ (3x− 2)

(3)8x3 − 1

2x− 1(4) (x3 − 1)÷ (x− 1)

Ex: Simplify

(1)25x12

5x4(2) (15x3 − 5x2 + x+ 5)÷ (5x)

(3) (8x4y5 − 10x4y3 + 12x2y3)÷ (4x3y2) (4) (x2 − 14x+ 24)÷ (x− 2)

(5) (4− 5x− x2 + 6x3)÷ (3x− 2) (6) (6x4 + 5x3 + 3x− 5)÷ (3x2 − 2x)

• 7.1 : Radical expressions and functions

Notice: 52 = 25 = (−5)2 , by√

25 we mean the positive value i.e.√

25 =√

52 = 5

Ex: Evaluate

(1)√

81 (2) −√

9 (3)

√4

49(4)√

36 + 64 (5)√

36 +√

64

The square root function

(a) Evaluate: Find f(3) given f(x) =√

2x− 2

(b) Graph: f(x) =√x and compare to f(x) = x2

Domain of a square root function

Ex: f(x) =√

3x+ 12 this has to be positive so set up inequality!

Square roots and cube roots

20

Square root:√

(a2) = |a| look at graphs of f(x) =√

(x2) and f(x) = |x|

(1)√

(−6)2 (2)√

(x+ 5)2 (3)√

25x6 (4)√x2 − 4x+ 4

Cube root: 3√

(a3) = a look at graphs of f(x) = 3√

(x3) and f(x) = x

(1) 3√

64 (2) 3√−27y3 (3)

3√

8x6

• 7.2 : Rational exponents

Ex:(

713

)3= 7

13·3 = 71 = 7

Definition: n√a = a

1n these are two different ways of writing the same thing

Ex: Compute

(1) 6412 =√

64 = 8 (2) (−125)13 = 3√−125 = 5

From the definition, ( n√a)

m=(a

1n

)m= a

1n·m = a

mn and so

(n√a)m

= amn also different ways of writing the same thing

Also from the definition, amn = am· 1

n = (am)1n = n

√am and so

amn = n

√am also different ways of writing the same thing

How to... simplify rational exponents and radicals

(1) Write out your number as a prime factor to some power

(2) Multiply the exponents using the appropriate “rule” of exponents i.e. (bm)n = bm·n

Ex: Compute

(1) 100023 (2) 16

32 (3) −32

35 (4) 27

23

All of the “rules” of exponents can also be used on fractional exponents

Recall: bmbn = bm+n Now:(a

kl

)(a

mn

)= a

kl+m

n

Recall: a−n =1

anNow: a−

mn =

1

amn

Also (bm)n = bm·n ,bm

bn= bm−n , etc.

Ex: Simplify

21

(1)10√x5 =

(x5) 1

10 = x510 = x

12 (2)

3√

27x15

(3) 4√x6y2 (4)

√x · 3√x (5) 3

√√x

• 7.3 : Multiplying and simplifying radical expressions

Ex: (√

4)(√

9) = 2 · 3 = 6 and also (√

4)(√

9) =√

36 = 6

Yes, this always works! Why...?? I’m glad you asked

(√

4)(√

9) = 412 · 9

12 = (4 · 9)

12 = (36)

12 = 6

The product “rule” for radicals√a√b =√a · b

Ex: Multiply

(1)√

3√

7 (2)√x+ 7

√x− 7 (3) 3

√4 3√

3

Simplifying radical expressions

How to...

(1) re-write as product of factors

(2) Take the root of appropriate factors

Ex: Simplify

(1)√

75 (2) 3√

54 (3) 5√

64 (4)√

500xy2

Simplifying a radical function

Ex: f(x) =√

2x2 + 4x+ 2 and g(x) =√

2|x+ 1| , look at graphs

This is one way to see why it is necessary to take the absolute value.

Notice: The book says “Assume that all variables in a radicand represent positive real num-bers,” however, we are going to learn the correct way to do this

Radicals “rule”!n√an = a if “n” is odd and

n√an = |a| if “n” is even

Note: |a2| = a2 and similarly |an| = an so long as “n” is even i.e.4√x8 = 4

√(x2)4 = |x2| = x2

Ex: Simplify

(1)3√x3 (2)

4√x8 (3)

√4x7 (4)

√9x9

22

(5)√x5y13z7 (2) 3

√32x8y16 (3) 5

√64x3y7z29

Multiply then simplify

Ex: (1)√

15√

3 (2) 7 3√

4 · 6 3√

6 (3) 4√

8x3y2 4√

8x5y3

• 7.4 : Adding, subtracting, and dividing radical expressions

Add / Subtract

“Like radicals” should be thought of as “like terms”

Ex: Add/Sub

(1) 7√

2 + 8√

2 (2) 3√

5−5 3√

5−2 3√

5 (3)√

7 + 2x√

7−2√

7

(4) 8 6√

5x− 5 6√

5x+ 4 3√

5x

Add/sub radicals after simplification

Ex: Add/Sub

(1) 7√

18 + 5√

8 (2) 4√

27x− 8√

12x (3) 7√

3− 2√

5

(4) 2 3√

16− 4 3√

54 5 3√xy2 + 3

√8x4y5

• 7.5 : Multiplying with more than one term and rationalizing denominators

Ex: Multiply

(1)√

7(x+√

2) (2) 3√x( 3√

6− 3√x2) (3) (5

√2 + 2

√3)(5√

2− 2√

3)

(4) (√

3 +√

7)2

Rationalizing denominators with one term

Ex: Write without a radical in the denominator

(1)

√5√6

(2) 3

√7

25(3)

√3√7

(4) 3

√2

9

Rationalizing denominators with two terms

Ex: Rationalize

(1)8

3√

2 + 4(2)

2 +√

5√6−√

3(3)

h√x+ h−

√x

(4)15√6 + 1

(5)

√11−

√5√

11 +√

5(6)

3√x+√y

√y − 3

√x

• 7.6 : Radical Equations

Ex:√x = 9

Solving Radical Equations

23

Ex: Solve

(1)√

2x+ 3 = 5 (2)√x− 3+6 = 5 (3) x+

√26− 11x = 4

(4)√

3x+ 1−√x+ 4 = 1 (5) (3x− 1)

13 + 4 = 0

• 7.7 : Complex numbers

Imaginary numbers!

Definition: i =√−1 and so i2 = −1

Ex:√−25 =

√25(−1) =

√25√−1 = 5i

Notice: To take the square root of a negative number,

√−b =

√b(−1) =

√b√−1 = (

√b)(i)

General written convention: “ 2i√

5 ” and “ 4i ” and “ i√

7 ”

Ex: Write as an imaginary number using “ i ”

(1)√−9 (2)

√−3 (3)

√−80

Complex numbers

Complex numbers can be written in the form

a+ bi where a and b are real numbers

We call a the real part of the complex number

We call b the imaginary part of the complex number

Notice: The “Complex numbers” contain the “Real numbers”

Adding and subtracting complex numbers

How to... Treat imaginary numbers as “like terms” , add the coefficients of “ i ”

i.e. (a+ bi) + (c+ di) = (a) + (c) + (b)i+ (d)i = (a+ c) + (b+ d)i and

(a+ bi)− (c+ di) = (a) + (−c) + (b)i+ (−d)i = (a− c) + (b− d)i

Ex: Simplify

(1) (5− 11i) + (7 + 4i) (2) (−5 + 7i)− (−11− 6i)

(3) (5− 2i) + (3 + 3i) (4) (2 + 6i)− (12− 4i)

24

Multiply complex numbers

Ex: Multiply

(1) 4i(3− 5i) (2) (7− 3i)(−2− 5i) (3) 7i(2− 9i) (4) (5 + 4i)(6− 7i)

Conjugates

Ex: (4 + 7i)(4− 7i) = 65 also recall (3−√

2)(3 +√

2) = 9− 2 = 7

Just like multiplying square root conjugates gets rid of the square root, multiplying com-plex conjugates gets rid of the imaginary part of a complex number.

Ex: Divide

(1)7 + 4i

2− 5i(2)

5i− 4

3i(3)

6 + 2i

4− 3i(4)

3− 2i

4i

Simplifying powers of “ i ”

How to...

(1) Write as “ i2 ” to some power i.e. i7 = (i6)(i) = (i2)3(i)

(2) Then simplify using i2 = −1 i.e. i7 = (i6)(i) = (i2)3(i) = (−1)3(i) = (−1)(i) = −i

Ex: Simplify

(1) i12 (2) i39 (4) i50

• 8.1 : “Square root property” and “completing the square”

The square root property is

if u2 = d, then u =√d or u = −

√d i.e. u = ±

√d

Ex: Solve

(1) x2 = 9 (2) 3x2 = 18 (3) 2x2 − 7 = 0

(4) 9x2 + 25 = 0 (5) (x− 1)2 = 5

Completing the square

This method ALWAYS works to solve quadratic equations. We use completing the square toderive the quadratic formula (after which we can just use the quadratic formula).

Given a quadratic equation with leading coefficient one

x2 + bx+ c = 0

25

We complete the square by adding and subtracting ( b2)2 like so

x2 + bx+

(b

2

)2

−(b

2

)2

+ c = 0

After which the first three terms factor as a “perfect square” like so(x+

b

2

)2

−(b

2

)2

+ c = 0

We then solve for the “perfect square”(x+

b

2

)2

=

(b

2

)2

− c

and finally use the “square root property” to finish solving.

Ex: Solve by completing the square

(1) x2−6x+4 = 0 (2) 9x2−6x−4 = 0 (3) 2x2−x+6 = 0

(4) x2+4x−1 = 0 (5) 2x2+3x−4 = 0 (6) 3x2−9x+8 = 0

• 8.2 : The Quadratic Formula

We start with a quadratic equation ax2 + bx+ c = 0 and solve for “x” to get the

“quadratic formula” x =−b±

√b2 − 4ac

2a

This method ALWAYS works to solve quadratic equations.

Ex: Solve

(1) x2+8x+12 = 0 (2) 2x2 = −4x+5 (3) 2x−√x−10 = 0

(4) (x2−5)2 +3(x2−5) = 0 (5) 10x−2 +7x−1 +1 = 0 (6) 5x23 +11x

13 +2 = 0

• 8.4 : Equations quadratic in form

The essential step for these questions is to identify the correct substitution to make.

Ex: Solve

(1) x4−8x2−9 = 0 (2) x4−5x2+6 = 0 (3) 2x−√x−10 = 0

(4) (x2−5)2 +3(x2−5) = 0 (5) 10x−2 +7x−1 +1 = 0 (6) 5x23 +11x

13 +2 = 0

26

• 8.3 : Graphing quadratic functions (and the discriminant from section 8.2)

Mini Lesson on the Discriminant

Given a quadratic function f(x) = ax2 + bx+ c , the discriminant is

b2 − 4ac (part of the quadratic formula)

(1) If b2 − 4ac > 0 , then the function has two real solutions (draw graph).

(2) If b2 − 4ac = 0 , then the function has one real solution (draw graph).

(3)b2 − 4ac < 0 , then the function has two imaginary solutions (draw graph).

Ex: Compute the discriminant, determine number and type of solutions.

(1) f(x) = 3x2 + 4x− 5 (2) f(x) = 3x2 − 8x+ 7 (3) f(x) = 9x2 − 6x+ 1

The graph of a quadratic function f(x) = ax2 + bx+ c = 0, (a 6= 0) is called a parabola.

Properties of Quadratic Functions

(1) Draw parabola with a < 0 and another with a > 0, label axis of symmetry and vertex

(2) Graph y = x2 and y = 2x2 and y = 12x

2 and y = x2 + 3 and y = x2 − 3

Graphing quadratic equations of the form f(x) = a(x− h)2 + k

Givenf(x) = a(x− h)2 + k

f(x) has a vertex at (h, k) and axis of symmetry at x = h. If a > 0 opens upward, if a < 0opens downward

How to... graph, given f(x) = a(x− h)2 + k form

(1) Plot vertex at the point (h, k)note: f(h) = k, so (h, k) is really (h, f(h))

(2) Determine if parabola opens upward or downward

(3) If parabola crosses x-axis, find x-intercept(s) by solving f(x) = 0

(4) Compute f(0) to find y-intercept

(5) Draw a nice arcing curve from vertex

Ex: Graph

(1) f(x) = −2(x− 3)2 + 8 (2) f(x) = (x+ 3)2 + 1

Graphing of the form f(x) = ax2 + bx+ c

Start with

f(x) = ax2 + bx+ c = a(x2 + bax) + c

27

and complete the square to get...

f(x) = a(x+ b2a)2 + c− b2

4a

Similar to f(x) = a(x− h)2 + k ?? Yes!!

“h” = − b

2aand “k” = c− b2

4a

Since “h” is the x-value of the vertex, the vertex is the point (− b2a , f(− b

2a))

How to... graph, given f(x) = ax2 + bx+ c form

(1) Find vertex at (− b2a , f(− b

2a))

(2) Determine if parabola opens upward or downward

(3) If parabola crosses x-axis, find x-intercept(s) by solving f(x) = 0

(4) Compute f(0) to find y-intercept

(5) Draw a nice arcing curve from vertex

Ex: Graph

(1) f(x) = −x2 − 2x+ 1 (2) f(x) = x2 − 3x+ 2 (3) f(x) = x2 − x+ 3

• 8.5 : Polynomial and rational inequalities

Ex: x2 − 1 > 0 or x2 − 1 < 0, draw graphs and show solutions

Solving polynomial inequalities

How to...

(1) Express the inequality in the form f(x) > 0 or f(x) < 0

(2) Solve the equation f(x) = 0. The real solutions are boundary points.

(3) Plot the boundary points on a number line, dividing the number line into intervals.

(4) Choose a “test value” from each interval

(a) If the number is positive, f(x) > 0 for the entire interval

(b) If the number is negative, f(x) < 0 for the entire interval

(5) Write the solution set including intervals that satisfy the given inequality

Ex: Solve

(1) x2 − 7x+ 10 (2) 2x2 + x > 15 (3) 4x2 ≤ 1− 2x (use quadratic eqtn)

(4) x3 + x2 ≤ 4x+ 4 (5) x3 + 3x2 < x+ 3

Solving rational inequalities

Ex:1

x> 0 or

1

x< 0, draw graphs and show solutions

28

How to...

Exactly the same as solving polynomial inequalities, except we change step (2)...

(2) Since f(x) is a “rational function” it can be written f(x) =p(x)

q(x)

(a) Solve p(x) = 0, solutions are boundary points

(b) Solve q(x) = 0, these are also

Ex: Solve

(1)x+ 3

x− 7< 0 (2)

x− 5

x+ 2< 0 (3)

x+ 1

x+ 3≥ 2 (4)

2x

x+ 1≥ 1

• 9.2 : Function composition and inverse functions

Function Composition

Ex: f(x) = x2 + 1 and g(x) = 3x+ 5 then f(g(x)) is...

The composition of a function f with another function g is written as (f ◦ g)(x) where

(f ◦ g)(x) = f(g(x))

Ex: Find (f ◦ g)(x) and (g ◦ f)(x) for

(a) f(x) = 2x− 7 and g(x) = x+ 3 (b) f(x) =√x and g(x) = −x+ 3

(c) f(x) = 2x2−x+1 and g(x) = x+2 (d) f(x) =x+ 1

2and g(x) = −x+5

Inverse functions

Ex: f(x) = 2x and g(x) = 12x are inverse functions since

f(g(x)) = f(12x) = 2(12x) = x and g(f(x)) = g(2x) = 12(2x) = x

Definition: Given a function f(x) and a function g(x) that satisfies

f(g(x)) = x and g(f(x)) = x for all values of x

we call g(x) the inverse function of f(x). We use the notation “f−1(x)” to denote the“inverse of the function f(x).”

Using this notation, f(f−1(x)) = x and f−1(f(x)) = x

Important! f−1(x) does NOT mean1

f(x)as in negative exponents.

Ex: Verify that the following functions are inverse functions

29

(a) f(x) = 5x and g(x) =x

5(b) f(x) = 3x+ 2 and g(x) =

x− 2

3

How to... (find the inverse of a given function)

(1) Replace f(x) with y

(2) Interchange x and y

(3) Solve for y

(4) Replace y with f−1(x)

Ex: Find the inverse of

(a) f(x) = 7x− 5 (b) f(x) = 2x+ 7

One-to-one functions and the horizontal line test

The “horizontal line test” is a way to tell if a function is one-to-one. A function must beone-to-one in order to have an associated inverse function f−1.

Why??

If a function isn’t one-to-one then the inverse will not be a function!

Ex: Which functions have inverses?

(a) f(x) = x+ 1 (b) f(x) = x2

Graphs of inverse functions

Ex: Graph function and inverse

(a) f(x) = 7x− 5 (b) f(x) = 2x+ 7

• 9.1 : Exponential functions

An exponential function f(x), with base b is

f(x) = bx or y = bx where b > 0 and b 6= 1

NOT exponential functions: f(x) = x2 , f(x) = 1x , f(x) = (−1)x , f(x) = xx

Graphing exponential functions

Ex: Graph using table

(a) f(x) = 2x (b) f(x) = (12)x = 2−x

then also draw graphs for f(x) = 7x and f(x) = (34)x

then also draw graphs for f(x) = 3x and f(x) = 3x+1 (shifted to left one unit)

30

Compound interest

Formula for compound interest: After t years, the balance A of an account with principle Pand annual interest rate r (in decimal form)

(1) For n compounding periods per year: A = P(

1 +r

n

)nt(2) For continuously compounding: A = Pert

Ex: You want to invest $8000 for 6 years, which choice is better?

(a) One choice pays 7% per year, compounded monthly

(b) The other pays 6.85% per year, compounded continuously

• 9.3 : Logarithmic functions

Logarithms are the inverse functions for exponential functions!

i.e. If f(x) = logb x and g(x) = bx then

(f ◦ g)(x) = logb bx = x

and(g ◦ f)(x) = blogb x = x

Definition of logarithmic functions

In the equationy = logb(x)

the value y is the answer to the question “To what power must b be raised, in order to getx?” i.e. What value of y must satisfy the equation

by = x

In these two equations we need b > 0, b 6= 1.

Ex: Change between logarithmic and exponential notation

(1) 2 = log5 x means (2) 3 = logb 64 means (3) log3 7 = y means

(4) 122 = x means (5) b3 = 8 means (6) ey = 9 means

Ex: Evaluate

(1) log2 16 (2) log3 9 (3) log25 5

Some properties of logs

(1) logb b = 1 Why?? (b1 = b)

(2) logb 1 = 0 Why?? (b0 = 1)

31

Graphing logs

Ex: Graph f(x) = 2x and g(x) = log2 x using tables.

How?? Do a table for f(x) and switch input/output in table for g(x).

Properties of logarithmic functions

(1) The domain of f(x) = logb x is (0,∞), and the range of f(x) = logb x is (−∞,∞)

(2) Notice for f(x) = logb x we have f(1) = logb 1 = 0, so x-intercept is always 1. There isnever a y-intercept.

(3) If b > 1 then f(x) = logb x is increasing function

(4) If 0 < b < 1 then f(x) = logb x is decreasing function

(5) The graph as a vertical asymptote at the y-axis

Domain of logs

The domain of f(x) = logb x is (0,∞), why?? (look at exponential equation)

Ex: Find domain

(1) f(x) = log4(x+ 3) (2) f(x) = log2(x− 5)

Note: If a log does not have a base then it is assumed to have base 10 i.e. log(2x) is log10(2x)

Natural logarithms

A logarithm with a base of “e” is a natural logarithm and is denoted

f(x) = lnx i.e. f(x) = loge x = lnx

Ex: Evaluate

(1) ln 1 (2) ln e7 (3) eln 15

• 9.4 : Properties of logarithms

The properties of exponents correspond to the properties of logarithms as follows...

The product rule for exponents:

bm · bn = bm+n corresponds to

The product rule for logs:

logb(mn) = logbm+ logb n

Ex: Expand each logarithmic expression

(1) log4(7 · 5) (2) log(10x) (= 1 + log x)

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The quotient rule for exponents:

bm

bn= bm−n corresponds to

The quotient rule for logs:

logb

(mn

)= logbm− logb n

Ex: Expand each logarithmic expression

(1) log7

(19

x

)(2) ln

(e3

7

)The power rule for exponents:

(bm)n = bmn corresponds to

The power rule for logs:logb(m

p) = p logbm

Ex: Expand each logarithmic expression

(1) log5(74) (2) ln

√x (3) log(10x)

(4) logb(x2√y) (5) log6

(3√x

36y4

)Ex: Condense each logarithmic expression

(1) log4 2 + log4 32 (2) log(4x− 3)− log x

(Note that coefficient out front must be 1 before you can proceed)

(3) 12 log x+ 4 log(x− 1) (4) 3 ln(x+ 7)− lnx (3) 4 logb x− 2 logb 6− 1

2 logb y

Change of base property:

logbm =logam

loga b

Here we are changing from a base of “b” to a base of “a”.

Ex: (i) First change to a base of 10 (ii) then change to a natural logarithm (base e)

(1) log5 140 (2) log6 17

• 3.1 : Systems of linear equations in two variables

Solving systems of linear equations by graphing

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Ex: Is (3, 4) a solution to this system?{x+ y = 7x− y = −1

Ex: Is (−4, 3) a solution to this system?{x+ 2y = 2x− 2y = 6

Solving by graphing

How to...

(1) Graph ’em both.

(2) If they intersect, find the point!

(3) Check the point in both equations.

Ex: Solve

(1) {3x+ 2y = 62x+ y = −2

(2) {y = −3x+ 2y = 5x− 6

Systems with ∞’ly many and zero solutions

Ex: One solution

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−5. −4. −3. −2. −1. 1. 2. 3. 4.

−5.

−4.

−3.

−2.

−1.

1.

2.

3.

4.

0

f

g

Ex: Infinitely many solutions

−5. −4. −3. −2. −1. 1. 2. 3. 4.

−5.

−4.

−3.

−2.

−1.

1.

2.

3.

4.

0

hg

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Ex: No solution

−5. −4. −3. −2. −1. 1. 2. 3. 4.

−5.

−4.

−3.

−2.

−1.

1.

2.

3.

4.

0

gf

A system with no solution: {y = 2x− 1y = 2x+ 3

A system with infinitely many solutions:{2x+ y = 3

4x+ 2y = 6

Solving systems using substitution

How to...

(1) Solve either equation for one of the variables.

(2) Substitute the equation found in (1) into the other equation, in the end the equation onlyhas one variable.

(3) Solve for the variable, this is half your answer.

(4) Substitute the result from (3) into equation from step (1).

(5) Check the solution in both equations.

Ex: Solve using the substitution method

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(1) {y = x− 1

4x− 3y = 24

(2) {5x− 4y = 9x− 2y = −3

Recognizing (1) the “no solution” situation and (2) the “infinitely many solu-tions” situation.

Ex: Same slope? Then no solution or infinitely many solutions.

(1) Same slope but different line {y − 5x = 4

y = 5x− 1

(2) These are the same line {y = −2x+ 3

4x+ 2y = 6

Solving systems using addition method

How to...

(1) Re-write the equations in Ax+By = C format.

(2) Multiply by a constant (if necessary) so that the sum of the x-coefficients or y-coefficientsis zero.

(3) Add equations from step (2), the sum should have only one variable.

(4) Solve for the variable.

(5) Substitute the result from (4) back into one of the original equations to find the othervariable.

Ex: Solve using the addition method

(1) {y = x− 1

4x− 3y = 24

(2) {5x− 4y = 9x− 2y = −3

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• 3.2 : Application questions using systems of equations

How to...

(1) What are you being asked to find? Label your variables.

(2) Figure out how your variables are related and write two equations.

(3) Solve for one variable, substitute back to find the other variable.

(4) Make sure you answered the question, and check your answer.

Ex: “Fencing a waterfront lot”

Perimeter of rectangular lot is 1000ft. Front decorative fencing costs $25 / foot, fencing fortwo sides costs $5 / foot, no rear fencing along water. If the total cost of the fencing was$9500, what are the dimensions of the lot?

Table: Cost per foot × number of feet = total cost{25x+ 2(5y) = 9500

2x+ 2y = 1000

Ex: A chemist needs to mix an 18% acid solution with a 45% acid solution to obtain 12 litersof a 36% acid solution.

Table: Acid per liter × number of liters = total acid{18%x+ 45%y = (36%)12

x+ y = 12

Ex: Flying with the wind a dragon can travel 450 miles in 3 hours. When the dragon turnsaround and goes against the wind she can travel the same distance in 5 hours. Find theaverage speed of the dragon in still air, and the average speed of the wind.

Table: Rate × time = distance {3(x+ y) = 4505(x− y) = 450

Ex: You have two options to install as heating systems. The solar system costs $29,700 toinstall and costs $150 / year. The electric system costs $5,000 to install and costs $1100 /year. After how many years will the total cost of each system be the same.

Table: {y = R(x) = 29, 000 + 150xy = S(x) = 5000 + 1100x

• 4.4 : Linear inequalities in two variables

Ex: 2x− 3y ≥ 6

Graphing using test points

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How to...

(1) Replace inequality with equality and graph.

(2) Choose a test point that is not on the line.

(3) If the inequality is true, shade that side of the line, otherwise shade the other side.

Ex: Graph

(1) 2x− 3y ≥ 6 (2) −x+ 2y > 4

Graphing inequalities in the form y < mx+ b and y > mx+ b

Ex: Graph

(1) y > −2

3x (2) y ≥ −x+ 1 (3) y ≥ −3 (4) x < 2

• 10.1 : Distance and midpoint; circles

The Distance Formula uses the pythagorean theorem, with d2 = (x2 − x1)2 + (y2 − y1)2,to compute the distance between two points (x1, y1) and (x2, y2)

d =√

(x2 − x1)2 + (y2 − y1)2

Ex: Find the distance between the points

(1) (0, 1) and (0, 3) (2) (−1, 4) and (3,−3)

The Midpoint Formula computes the midpoint of a line segment connecting points (x1, y1)and (x2, y2)

midpoint is

(x1 + x2

2,y1 + y2

2

)A good way to see this formula is that

x1 + x22

is the average of x1 and x2.

Ex: Find the midpoint of the line segment connecting the points

(1) (0, 1) and (0, 3) (2) (1,−6) and (−8,−4)

Circles!

Definition: A circle of radius r centered at the point (h, k) is the set of all points a fixeddistance “ r ” from a single point (h, k).

The “standard” form of the equation of a circle of radius r centered at the point (h, k)

(x− h)2 + (x− k)2 = r2

This is the distance formula used with the center point (h, k) and a point (x, y) on the circle.

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Ex: Write the equation of a circle of radius r = 3 centered at the point (2, 3)

Ex: Graph

(1) (x− 2)2 + (y − 3)2 = 9 (2) (x+ 1)2 + (y + 3)2 = 1

The more general form of the equation of a circle can be turned into the above “standard”form by completing the square

x2 + y2 +Dx+ Ey + F = 0

Ex: Graph

(1) x2 + y2 + 4x− 6y − 23 = 0 (2) x2 + y2 + 4x− 4y − 1 = 0

• 10.2 : The ellipse!

Definition: An ellipse is the set of all points (x, y) the sum of whose distances from two fixedpoints P1 and P2 is constant. These fixed points are called the Foci and between them liesthe Center of an ellipse.

The “standard” form of the equation of an ellipse centered at the origin

x2

a2+y2

b2= 1

with x-intercepts {(a, 0), (−a, 0)} and y-intercepts {(0, b), (0,−b)}.

Ex: Graph

(1)x2

9+y2

4= 1 (2) 25x2 + 16y2 = 400

• 10.3 : The hyperbola

Definition: A Hyperbola is the set of all points (x, y) the difference of whose distances fromtwo fixed points P1 and P2 is constant. These fixed points P1 and P2 are called the Foci andbetween them lies the Center of an ellipse.

The “standard” form of the equation of a hyperbola centered at the origin

x2

a2− y2

b2= 1

with x-intercepts {(a, 0), (−a, 0)} and no y-intercepts.

Ex: Graph

(1)x2

9+y2

4= 1 (2) 25x2 + 16y2 = 400

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• 10.4 : The parabola; identify conic sections

Definition: A parabola is the set of all points (x, y) that are equidistant from a fixed line(the “directix”) and a fixed point (the “foci”)

• 10.5 : Systems of nonlinear equations in two variables

What do solutions look like?

Ex: A parabola intersecting a circle {y = x2

x2 + y2 = 1

Substitution method

Ex: Solve(A line intersecting a parabola) {

x2 = 2y + 103x− y = 9

Elimination method

Ex: Solve(Intersecting circles) {

4x2 + y2 = 13x2 + y2 = 10

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