#### Description

Grade 10 AP Mathematics Sasha Kheyson 1.0 - PREREQUISITE SKILLS ................................................................................................................... 3 1.1 - ORDERED PAIRS AND SOLUTIONS ............................................................................................. 3 1.2 - SOLVING LINEAR SYSTEMS GRAPHICALLY ........................................................................... 4 1.3 - SOLVING LINEAR SYSTEMS BY SUBSTITUTION .................................................................... 5 1.5 - SOLVING LINEAR SYSTEMS BY ELIMINATION ...................................................................... 5 1.6 - HOW TO SOLVE WORD EQUATIONS .......................................................................................... 6 1.7 - WORD PROBLEMS ............................................................................................................................ 6 INVESTMENT PROBLEM ............................................................................................................................. 6 MIXTURE PROBLEM .................................................................................................................................. 7 DISTANCE - SPEED - TIME PROBLEM ......................................................................................................... 7 RIVERBOAT CRUISE................................................................................................................................... 8 2.1 - THE PYTHAGOREAN THEOREM.................................................................................................. 9 2.1 - EQUATION OF A CIRCLE................................................................................................................ 9 2.2 - EQUATION OF A LINE ..................................................................................................................... 9 PARALLEL LINE ....................................................................................................................................... 10 PERPENDICULAR LINES ........................................................................................................................... 10 2.3 - MIDPOINT OF A LINE SEGMENT ............................................................................................... 10 2.4 - TRIANGLES....................................................................................................................................... 11 MEDIANS ................................................................................................................................................. 11 RIGHT BISECTORS ................................................................................................................................... 11 ALTITUDES .............................................................................................................................................. 11 CENTROID ............................................................................................................................................... 12 CIRCUMCENTRE ...................................................................................................................................... 12 ORTHOCENTRE ........................................................................................................................................ 12 CHORDS .................................................................................................................................................. 12 2.5 - DISTANCE FROM A POINT TO A LINE...................................................................................... 13 SHORTCUT ............................................................................................................................................... 13 3.1 - POLYNOMIALS ................................................................................................................................ 14 CLASSIFYING POLYNOMIALS ................................................................................................................... 14 EXPONENT RULES ................................................................................................................................... 14 3.4 - COMMON FACTORS....................................................................................................................... 14 FACTORING FULLY .................................................................................................................................. 14 FACTORING COMMON BINOMIAL FACTORS............................................................................................. 14 FACTORING BY GROUPING ...................................................................................................................... 15 3.5 - FACTORING TRINOMIALS ........................................................................................................... 15 3.5 - FACTORING TRINOMIALS WHEN A ≠ 1 ............................................................................................. 15 3.7 - FACTORING SPECIAL QUADRATICS ........................................................................................ 16 DIFFERENCE OF SQUARES........................................................................................................................ 16 PERFECT BINOMIAL SQUARES ................................................................................................................. 16

1

2

1.0 - Prerequisite Skills Simplify: 2x2 - 4x(x + 3) = 2x2 - 4x2 - 12x = -2x2 - 12x

--> Distributing Properties. --> Collecting Like Terms.

Solve: x+2=7 x=7-2 x=5

--> Find the value of variable(s) that makes the equation true.

Check:

--> Substitute the value of the variable(s) in the original equation and check for equality.

Left Side = x + 2 =5+2 =7

Right Side = 7

Left Side = Right Side

1.1 - Ordered Pairs and Solutions The equation 2x + y = 4 can be graphed in three different ways: 1)

Using a table of values:

y = -2x + 4 2x + y = 4

2)

x -1 0 1

y 6 4 2

Table of Values

Ordered Pairs

Using x and y intercepts: x intercept: let y = 0, then 2x = 4 x=2

3)

(-1, 6) (0, 4) (1, 2)

y intercept: let x = 0, then y=4

(2, 0) (0, 4)

Using the slope-y intercept equation: y = mx + b 2x + y = 4 y = -2x + 4

y intercept: 4

Slope: -2/1

3

1.2 - Solving Linear Systems Graphically An equation with one variable has one real solution. An equation with two variables can have an infinite number of solutions. x+y=5 (x, y): ordered pair (5, 1); (6, -1); (1.5, 3.5)... A linear system is two or more linear equations on the same Cartesian plane. Solving a linear system means finding the point of intersection of the two lines. There are three methods to solve a linear system: 1. Graphing 2. Substitution 3. Elimination As well, there are three types of solutions: 1. One solution: (x, y): 2. No solutions:

3. Infinite number of solutions:

For example: x-y=4 (5, 1); (4, 0) x+y=2 (1, 1); (2, 0)

The point of intersection is at (3, -1).

4

1.3 - Solving Linear Systems by Substitution 1. Solve one equation for one of its variables (either x or y). 2. Substitute the known variable into the other equation and solve for the remaining variable. 3. Substitute the variable found in the second equation back into the first equation, and solve for the remaining variable. 4. Check both equations. 2x + y = 2 3x + 2y = 1

--> Solve for the variable y: y = 2 - 2x --> Substitute the above equation into the second one: 3x + 2(2 - 2x) = 1

3x + 4 - 4x = 1 -x = -3 x=3 --> Solve the equation for the variable x. 2(3) + y = 2 --> Replace the x value back into the first equation. 6+y=2 y = -4 --> Solve the equation for the variable y. (x, y) = (3, 4).

1.5 - Solving Linear Systems by Elimination The objective of elimination is to add linear equations together (vertically), thereby eliminating one of the two variables. x+y=4 +x -y=2 2x = 6 x =3

--> Line up the two equations (the x's, y's and constants are aligned). --> Add both equations together. One variable should by eliminated. --> Solve for the remaining variable.

x+y=4 3+y=4 y=1 (x, y) = (3, 1)

--> Go back to one of the original equations, and substitute the known variable there. --> Solve for the last variable. --> The point of intersection.

************************************************************************ x - 2y = 7 3x + 4y = 1

--> Adding these two equations will not eliminate any variables. --> Multiply the first equations by 2, which will eliminate the y variable.

2x - 4y = 14 3x + 4y = 1 5x = 15 x =5

--> The two equations are added, and the y variable disappears. --> Solve for x, then solve for y.

5

1.6 - How to Solve Word Equations 1. 2. 3. 4. 5. 6. 7. 8. 9.

Read the question. Read the question. Again. What am I being asked to find out? What relevant information am I given (i.e. chart or graph)? Introduce variables: "Let x represent..." Create equations. Solve linear systems (algebraically). Check the solution for reasonableness (i.e. no negative distances). Concluding statement (include units of measurement).

1.7 - Word Problems Investment Problem Bob had \$10 000 to invest. He invested part of it at 4% per annum (per year), and the rest at 5% per annum. If the total interest was \$440, how much did he invest at each rate? Let x represent the amount he invested at 4% per annum. Let y represent the amount he invested at 5% per annum. x + y = 10 000 0.04x + 0.05y = 440

--> First equation (total money invested). --> Second equation (total money earned).

x = 10 000 - y 0.04(10 000 - y) + 0.05y = 440 400 - 0.04y + 0.05y = 440 0.01y = 40 y = 4 000

--> Rearrange the equation as x = ... . --> Substitute x into the second equation. --> Expand the equation.

x + 4 000 = 10 000 x = 6 000

--> Substitute y back into the first equation. --> Solve for x.

--> Solve for y.

Bob invested \$6 000 at 0.04% per annum, and \$4 000 at 0.05% per annum. ************************************************************************

6

Mixture Problem Mr. Stewart needs to make 10L of a 42% acid solution. The problem is that the acid solutions are only available in 30% and 50% by volume concentrates. How many litres of each solution must be mixed to make the 42% acid solution?

30% acid solution 50% acid solution Required % solution

Volume (L) x y 10

% Concentration 0.3 0.5 0.42

Final Solution 0.3x 0.5y 4.2

Let x represent the required volume of the 30% acid solution. Let y represent the required volume of the 50% acid solution. x + y = 10 0.3x + 0.5y = 4.2

--> The first equation (total number of litres). --> The second equation (% concentration).

x = 10 - y 3x + 5y = 42 3(10 - y) + 5y = 42 30 - 3y + 5y = 42 2y = 12 y=6

--> Rearrange the equation as x = ... . --> Multiply the equation by 10 to get rid of the decimal. --> Substitute x into the second equation.

x + 6 = 10 x=4

--> Substitute y back into the first equation. --> Solve for x.

--> Solve for y.

Mr. Stewart needs 4L of 30% acid solution, and 6L of 50% acid solution. ************************************************************************ Distance - Speed - Time Problem Henry drove 500km from Toronto to Ottawa in 5.5 hours. He drove part of the distance on Hwy 401 at 100km/h, and the rest of the way on Hwy 7, at 80km/h. How far did he drive on each highway? Hwy 401 Hwy 7 Total

Distance (km) x y 500

Speed (km/h) 100 80 N/A

Time (hours) x ÷ 100 y ÷ 80 5.5

Let x represent the distance he drove on Hwy 401. Let y represent the distance he drove on Hwy 7.

7

x + y = 500 x ÷ 100 + y ÷ 80 = 5.5 x = 500 - y 8x + 10y = 4400 8(500 - y) + 10y = 4400 4000 -8y + 10y = 4400 2y = 400 y = 200

--> The first equation (total distance). --> The second equation (total time at each speed). --> Rearrange the equation as x = ... . --> Multiply the equation by 800 to get rid of the decimal. --> Substitute x into the second equation.

x + 200 = 500 x = 300

--> Substitute y back into the first equation. --> Solve for x.

--> Solve for y.

Henry drove 300km on Hwy 401, and 200km on Hwy 7. ************************************************************************ Riverboat Cruise A boat took 2 hours to travel 24km downstream with the current, and 3 hours to make the return trip, going against the current. Find the speed of the boat in still water, and the speed of the current. Downstream Upstream

Distance (km) 24 24

Speed (km/h) x+y x-y

Time (hours) 24 ÷ (x + y) 24 ÷ (x - y)

Let x represent the speed of the boat. Let y represent the speed of the current. 24 ÷ (x + y) = 2 24 ÷ (x - y) = 2

--> The first equation (downstream). --> The second equation (upstream).

12 = x + y 8=x-y

--> Rearrange the first equation as x + y = ... . --> Rearrange the second equation as x - y = ... .

x -y=8 x + y = 12 2x = 20 x = 10

--> Use elimination to solve for x.

24 ÷ (10 + y) = 2 24 = 2(y + 10) 12 = y + 10 y=2

--> Solve for x. --> Substitute y back into the first equation.

--> Solve for y.

The speed of the boat is 10km/h, and the speed of the current is 2km/h.

8

2.1 - The Pythagorean Theorem This equation looks at how to find the distance of a hypotenuse of a right-angle triangle on the Cartesian plane. (AB)2 = (AC)2 + (BC)2 (AB)2 = 62 + 62 (AB)2 = 72 (AB) ≈ 8.5 A (4, 2) (x1 y1)

B (10, 8) (x2 y2)

(AB)2 = (x2 - x1)2 + (y2 - y1)2 √ (AB)2 = √ [(x2 - x1)2 + (y2 - y1)2] AB = √ [(x2 - x1)2 + (y2 - y1)2]

C (10, 2) (x2 y1)

2.1 - Equation of a Circle This is the general equation of a circle whose centre is the origin (0, 0). AB = √ [(x2 - x1)2 + (y2 - y1)2] AB = √ [(x - 0)2 + (y - 0)2] AB = √ (x2 + y2) Radius = √ (x2 + y2) Radius2 = (x2 + y2) r2 = x2 + y2

2.2 - Equation of a Line

• •

Equations can be represented in two forms: Slope y intercept form: y = mx + b, where m is the slope, and b is the y intercept. Standard form: Ax + By + C = 0, where A, B and C are integers, and A > 0.

• •

To find the equation of a line, you need two pieces of information: The slope (m). A point the line goes through (x, y). The point-slope formula is: y - y1 = m(x - x1). Find the equation of the line that passes through the points A (3, 4) and B (-1, 2).

m = (y1 - y2) ÷ (x1 - x2) = (4 - 2) ÷ (3 - -1) =2÷4 =1÷2

y - y1 = m(x - x1) y - 4 = 1 / 2(x - 3) 2y - 8 = x - 3 x - 2y + 5 = 0

The equation of the line that passes trough the points A and B is x - 2y + 5 = 0.

************************************************************************

9

Parallel Line Two lines are parallel when they have the same slope: m1 = m2. The two lines will never intersect. y = 3x + 12 and y = 3x - 2 are parallel because their slopes (m) are the same: 3/1. ************************************************************************ Perpendicular Lines Two lines are perpendicular when their slopes are negative reciprocals: m1 x m2 = -1. Two lines will intersect at a right (90°) angle. y = 2x + 6 and y = -1 / 2x - 9 are perpendicular because their slopes (m) are negative reciprocals.

2.3 - Midpoint of a Line Segment Midpoint AB = (average of x's, average of y's), given A (x1, y1) and B (x2, y2). Midpoint AB = [(x1 + x2) ÷ 2, (y1 + y2) ÷ 2] Given the endpoints of a line segment are A(-2, -3) and B(4, 7), find the midpoint. MAB = [(x1 + x2) ÷ 2, (y1 + y2) ÷ 2] MAB = [(-2 + 4) ÷ 2, (-3 + 7) ÷ 2] MAB = [(2 ÷ 2), (4 ÷ 2)] MAB = (1, 2) The midpoint of AB is M(1, 2). Given the endpoint A(-2, -3) and the midpoint of CD is M(1, 2), find the coordinates of the other endpoint, D. MCD = [(x1 + x2) ÷ 2, (y1 + y2) ÷ 2] M(1, 2) = [(-2 + x2) ÷ 2, (-3 + y2) ÷ 2] 2 = (-3 + y2) ÷ 2 1 = (-2 + x2) ÷ 2 4 = (-3 + y2) 2 = (-2 + x2) y2 = 7 x2 = 4

--> Substitute the midpoint of the line. --> Make the x of the midpoint (1) equal the x in the equation. Do the same for y.

The endpoint D is located at (4, 7).

10

2.4 - Triangles Medians A median is a line segment that joins the vertex of a triangle to the midpoint of the opposite side. A (x1, y1) m

B (x2, y2)

C (x3, y3)

1. Find the midpoint AC: [(x1 + x2) ÷ 2, (y1 + y2) ÷ 2]. 2. Find the slope of the median from B: (y1 - y2) ÷ (x1 - x2). 3. Use the point-slope formula with the point B: y - y1 = m(x - x1). Right Bisectors A right bisector is a line that passes through the midpoint of a line segment at a right angle. D (x1, y1)

E (x2, y2) 1. 2. 3. 4.

F (x3, y3)

Find the midpoint EF: [(x1 + x2) ÷ 2, (y1 + y2) ÷ 2]. Find the slope of EF: (y1 - y2) ÷ (x1 - x2). Find the negative reciprocal (perpendicular slope) of EF: m = -1 ÷ m. Use the point-slope formula with the midpoint of EF: y - y1 = m(x - x1). Altitudes

An altitude is a line segment that joins a vertex to an opposite side at a right angle. G (x1, y1)

H (x2, y2)

I (x3, y3)

1. Find the slope of HI: (y1 - y2) ÷ (x1 - x2). 2. Find the negative reciprocal (perpendicular slope) of HI: m  -1 ÷ m. 3. Use the point-slope formula with the perpendicular slope and point G.

11

Centroid The centroid is the point of intersection of all three medians of a triangle.

1. Find the equation of all three medians. 2. Find the point of intersection of any two medians. 3. Use the third median to verify if the point of intersection is correct. Circumcentre The circumcentre is the point of intersection of all three right bisectors of a triangle.

1. Find the equation of all three right bisectors. 2. Find the point of intersection of any two right bisectors. 3. Use the third right bisector to verify if the point of intersection is correct. Orthocentre The orthocentre is the intersection of all three altitudes of a triangle.

4. Find the equation of all three altitudes. 5. Find the point of intersection of any two altitudes. 6. Use the third altitude to verify if the point of intersection is correct. Chords A chord is a line segment joining two points on the circumference of a circle. If you draw all three right bisectors, altitudes and medians on a right triangle, their centroid, orthocentre and circumcentre will be co-linear.

12

2.5 - Distance from a Point to a Line The shortest distance a point is from a line is when it is perpendicular. Find the shortest distance (d) from the point B(10, -3) to the line y = 2x + 3.

B (10, -3)

y = 2x + 3 m=2÷1 m= -1 ÷ 2 B (10, -3)

--> The equation of the line. --> The slope of the line. --> The perpendicular slope. --> The point.

y - y1 = m(x - x1) y + 3 = -1 ÷ 2(x - 10) 2y + 6 = 10 - x y = -x ÷ 2 + 2

--> Point-slope formula. --> Point B is substituted in. --> Simplifying. --> Equation of line segment.

2x + 3 = -x ÷ 2 + 2 4x + 6 = -x + 4 5x = -2 x = -2 ÷ 5

--> Substitute the first equation of line into the second. --> Simplify. --> Solve for x.

y = 2x + 3 y = 2(-2 ÷ 5) + 3 y = -4 ÷ 5 + 3 y = 11 ÷ 5

--> Substitute the value of x into one of the equations. --> Simplify. --> Solve for y.

The point of intersection is P(-2/5, 11/5). The distance between two points can be found using the following equation: PB = √ [(x2 - x1)2 + (y2 - y1)2] PB = √ [(-2 ÷ 5 - 10)2 + (11 ÷ 5 + 3)2] PB = √ (108.16 + 27.04) PB = √ 135.2 PB ≈ 11.6 Shortcut y = 2x + 3 2x - y + 3 = 0 A = 2, B = -1, C = 3

--> The original equation of line. --> Change the equation into standard form: Ax + By + C = 0. --> The values of A, B and C, respectively.

d = [|Ax1 + By1 + C|] ÷ [√ (A2 + B2)] --> The shortcut equation. d = [|(2 x 10) + (-1 x -3) + (3)|] ÷ √ [22 + (-1)2] d = |20 + 3 + 3| ÷ √ (4 + 1) d = |26| ÷ √ 5 d ≈ 11.6 --> The distance from B (10, -3) to the line y = 2x + 3.

13

3.1 - Polynomials Classifying Polynomials There are two ways to classify polynomials: 1. Number of terms (monomial, binomial, trinomial, polynomial). 2. Degree of a monomial (-2xy2z3 = degree 6) or of a polynomial (5x2 + xy + 2x + 5 = degree 2). As well, a polynomial cannot have a negative exponent or a negative denominator. Exponent Rules • • • • • • • • •

Multiplication (am x an = am + n): if the bases are the same, add the exponents. Division (am ÷ an = am - n): if the bases are the same, subtract the exponents. Power of a Power [(am)n = amn]: multiply the exponents. Power of a Product [(a x b)m = ambm]: apply exponent to all factors. Power of a Quotient [(a ÷ b)m = am ÷ bm]: apply exponent to the numerator and the denominator. Exponents of Zero (x0 = 1): any base raised to the exponent 0 is equal to 1. Negative Exponents (a-m = 1 ÷ am): remove the negative sign and out the base and exponent as the denominator with a numerator of 1. Rational Exponents (a1÷m = m√a): index: denominator (m), radicand: base (a). Rational Exponents [an÷m = (m√a)n]: index: denominator (m), radicand: base (a), exponent: numerator (n).

3.4 - Common Factors Expanding means to distribute polynomials. Simplifying means to collect like terms and rearrange them in descending order. The opposite of expanding and simplifying is factoring. Factoring means to break an equation into two or more factors. Factoring Fully 12 = 2x2x3 3x - 6 = 3(x - 2) 12x2 - 6x = 2x(6x - 3)

--> To factor an equation fully, the factor removed must be the greatest common factor (GCF).

Factoring Common Binomial Factors 3x(x + 1) + 4y(x + 1) = (x + 1)(3x + 4y)

--> The binomial (x + 1) is common to both factors. --> It can be removed as shown.

14

Factoring by Grouping ac + bd + bc + ad = ac + ad + bc + bd = a(c + d) + b (c + d) = (c + d)(a + b)

--> Original equation. --> Rearrange the terms so that factorable terms are together. --> Factor out a common factor from each pair of terms. --> Factor out the common binomial (c + d).

3.5 - Factoring Trinomials (x + 2)(x - 3) = x2 + 2x - 3x - 6 = x2 - x - 6

--> Expand. --> Use FOIL. --> Simplify.

x2 - x - 6 (x + 2)(x - 3)

--> Factor. --> The factored product.

(a + b)(a + c) --> General equation. 2 = a + ac + ab + bc --> Expanded equation. --> Simplified equation. Here, the second term is a(b + c): the sum = a2 + a(b + c) + bc of b and c multiplied by a. The third term is the product of b and c. Normally, the first term has a degree of 2, the second term has a constant multiplied by a variable, and the last term is a constant. The sum of the products of the last term must equal the sum of the numerical coefficient of the second term.

3.5 - Factoring Trinomials when a ≠ 1 (6x2 + 23x + 20) 3x 2x

--> The original equation. 1 20

2 10

4 5

5 4

10 2

20 1

This method of factoring is called the cross-multiplication method. When you have removed the GCF from the equation, look at the a coefficient, in this case 6. Factor this coefficient into any two factors: 2 and 3 or 1 and 6. at the last constant in the trinomial, in this case 40. List all the factors of the constant, in a logical order so that you don't miss any. Next, look at the b term: 23x. This is the sum you are trying to obtain. First, multiply the 3x by 20, and the 2x by 1: you take the two factors of the a coefficient and cross-multiply them by the two factors of the c constant, one pair at a time. Then add the result together. (3x)(20) + (2x)(1) = 60x + 2x = 62x. This is not the b term. Move on to the next pair of numbers. (3x)(10) + (2x)(2) = 30x + 4x = 34x --> 34x is not equal to 23x. (3x)(5) + (2x)(4) = 15x + 8x = 23x --> 23x is equal to 23x. Now, go back to the table. Put brackets around the correct set of factors. Ignoring the (3x 1 2 4) 5 2 1 (2x 20 10 5) 4 10 20 numbers in between, look at the two lines. Your factors are (3x + 4) and (2x + 5).

15

3.7 - Factoring Special Quadratics Difference of Squares (x + 5)(x - 5) = x2 + 5x - 5x - 25 = x2 -25

--> The original equation. --> Expand. --> Simplify, becomes a difference of squares.

When you multiply two binomials together that look the same but have a different sign between them, the two middle signs will always cancel each other out, leaving the first and the last term. This product is called a difference of squares, because if you look at it, you have two squares [x2 and 25, where x2 = (x)(x) and 25 = (5)(5)], separated by a subtraction. There is a shortcut to expand a difference of squares. First, square the first term of one of the binomials (it doesn't matter which one). Then, square the second term of the binomial and add a negative sign in front. For example: (x + 11)(x - 11) = x2 - 121

x2 - 121 = (x + 11)(x - 11)

If you see a difference of squares, you can also factor it back into the same form, as a product of two binomials with opposite signs. Perfect Binomial Squares (x + 5)2 = x2 + 5x + 5x + 25 = x2 + 10x + 25

--> The original equation. --> Expand. --> Simplify, becomes a perfect binomial square.

When you multiply two binomials together that are exactly the same (a binomial squared), the result is a perfect binomial square. There is a shortcut to expand a perfect binomial square. First, square the first term of the binomial. Then, multiply both terms together and double the product. Lastly, square the last term. It is important to note that the first and last term will always be positive, because they are perfect squares. (x + 11)2 = x2 + 22x + 121

x2 + 22x + 121 = (x + 11)(x + 11)

If you see a perfect binomial square, you can also factor it back into the same form, as a binomial squared.

16

3.8 - Rational Expressions Simplifying Rational Expressions A rational expression is the quotient of polynomials [x ÷ 3, x ÷ (x - 2)]. If a variable appears in the denominator, there is a restriction for the variable: the denominator cannot be equal to 0. [x ÷ (x - 1)], where x cannot be equal to 1.

1. 2. 3. 4.

There are several steps to simplify a rational expression: Factor the numerator. Factor the denominator. State the restrictions on the variable. Reduce (only when multiplying between terms).

(x2 - x) ÷ x --> The original equation. = x(x - 1) ÷ x --> Factor the numerator. = x - 1, x cannot be 1 --> Reduce the numerator and the denominator, state restriction. (x2 + 3x + 2) ÷ (x2 - x - 2) = (x + 1)(x + 2) ÷ (x - 2)(x + 1) x cannot be equal to 2, -1 = (x + 2) ÷ (x - 2)

--> The original equation. --> Factor the numerator and the denominator. --> State restrictions on the denominator. --> Reduce the numerator and the denominator.

Multiplying Rational Expressions 1. 2. 3. 4.

Factor the numerators of all the terms. Factor the denominators of all the terms. State the restrictions on the denominators of each term. Reduce the numerators and denominators. Dividing Rational Expressions

1. 2. 3. 4. 5. 6.

Factor the numerators of all the terms. Factor the denominators of all the terms. State the restrictions on the denominators of each term. Change the divisions into multiplications. State the restrictions again. Reduce the numerators and denominators.

[(x2 - 4) ÷ (2x2 - 5x + 2)] ÷ [(x2 - 2x) ÷ (2x2 - 3x - 2)] = [(x + 2)(x - 2) ÷ (2x - 1)(x - 2)] ÷ [(x)(x - 2)÷ (2x + 1)(x - 2)] x cannot be equal to 1/2, 2, -1/2 = [(x + 2)(x - 2) ÷ (2x - 1)(x - 2)] x [(2x + 1)(x - 2) ÷ (x)(x - 2)] x cannot be equal to 0 = [(1) ÷ (2x - 1)] ÷ [(2x + 1)(x - 2) ÷ (x)] = (2x + 1)(x - 2) ÷ (x)(2x - 1)

--> Original equation. --> Steps 1 and 2. --> Step 3. --> Step 4. --> Step 5. --> Step 6. --> Step 6.

17

Adding and Subtracting Rational Expressions 4 / (3y - 3) + 3 / (5y + 5) = 4 / 3(y - 1) + 3 / 5(y + 1) = [4 x 5 x (y + 1)] + [3 x 3 x (y - 1)] / [3 x (y + 1) x 5 x (y + 1)] = [20y + 20 + 9y - 9] / [15(y + 1)(y - 1)] = (29y + 11) / [15(y + 1)(y - 1)] 4 / (x2 + 5x + 6) - 5 / (x2 - x - 12) = 4 / (x + 2)(x + 3) - 5 / (x + 3)(x - 4) = [4(x - 4) - 5(x + 2)] / (x + 2)(x + 3)(x - 4) = (4x - 16 - 5x -10) / (x + 2)(x + 3)(x - 4) = -x - 26 / (x + 2)(x + 3)(x - 4)

--> The original equation. --> Factor the denominator. --> Common denominator. --> Expand. --> Simplify.

--> The original equation. --> Factor the denominator. --> Common denominator. --> Expand. --> Simplify.

3.9 - Solving First Degree Inequalities First-degree inequalities in one variable are solved by using the same principles as those in the Equality Property for Equations, except when multiplying or dividing by a negative number. There are 5 inequality signs: > ... greater than < ... less than

≥ ... greater than or equal to ≤ ... less than or equal to

Original Inequality 9>6 9>6 9>6 9>6 9>6 9>6

≠ ... not equal to

Operation Add 3 Subtract 3 Multiply by 3 Multiply by -3 Divide by 3 Divide by -3

Resulting Inequality 9+3>6+3 9-3>6-3 9x3>6x3 9 x -3 < 6 x -3 9/3>6/3 9/3 A radical is only simplified when all the perfect squares/cubes are removed from the radicand.

18

Rationalizing a Denominator 1 / √2 = (1 / √2) x 1 = (1 / √2) x (√2/√2) = √2 / 2

--> Because a radical cannot be part of a denominator, it must be removed by rationalizing the fraction. This is done by multiplying the entire fraction by 1, in the form of the radical in the denominator. Adding Radicals

(1 / √2) + (1 / √3) = (√2 / 2) + (√3 / 3) = (3√2 + 2√3) / 6

--> Like adding any fractions with radicals, the radical must first be removed from the denominator by rationalizing it. Then, a denominator is achieved, and the fractions can be simplified. Conjugates

1 / (√2 + 2) = 1 / (√2 + 2) x 1 = 1 / (√2 + 2) x (√2 - 2) / (√2 - 2) = (√2 - 2) / 2 - 4 = (√2 - 2) / -2

--> If the radical in the denominator is part of a binomial, a conjugate is required to eliminate it. Like a difference of squares, the outside and the inside terms eliminate when multiplied together. A conjugate works based on the same principle.

4.1 - Quadratic Relations and Functions A relation is a set of ordered pairs, such as (x, y). The domain is the set of all the first elements of the ordered pairs in a relation. The range is the set of all the second elements of the ordered pairs in a relation. Relation: Domain: Range:

(1, 2), (3, 4), (5, 6),(7, 8) 1, 3, 5, 7: 2, 4, 6, 8:

{1 ≤ x ≤ 7, x ε I} {2 ≤ y ≤ 8, x ε I}

A function is a set of ordered pairs, where for every value of x, there is one, and only one, value of y. To determine if a relation is a function or not, use the "vertical line test". Alternatively, you can compare all the x co-ordinates in the relation. If they do repeat, the relation is not a function. Relation: (-2, 3),(-1, 2),(0, 1),(1, 2),(2, 3) Domain: {-2 ≤ x ≤ 2, x ε I} Range: {1 ≤ y ≤ 3, y ε I} Function: yes

19

4.2 - The Basic Parabola A quadratic function is a function defined by the quadratic equation in the form y = ax + bx + c. The graph of a quadratic function is the parabola: y = x2. It is called the basic parabola. 2

x = y2

x -2 -1 0 1 2

---> The vertex is the turning point of the parabola [co-ordinate (x, y)]. The axis of symmetry is the reflection line that divides the parabola vertically down the middle (described as an equation of a line). Domain: {x ε R} Range: {0 ≤ y, y ε R}

y 4 1 0 1 4

The value in front of the 'x' in (y = x2) determines the shape of the parabola (how wide it opens). Adding or subtracting a constant (y = x2 + 3) does not affect the shape of the parabola, but rather moves it up or down. (y = x2 - 2) and (y = x2 + 7) have the same shape, but are in different locations on the y axis. They are said to be congruent. The polarity of 'a' in (y = ax2 + k) will determine the direction of the opening of the parabola. If 'a' is positive, the parabola will open upward; if 'a' is negative, the parabola will open downward. The value of 'k' will determine the y coordinate of the parabola's vertex. Since the vertex always lies on the axis of symmetry, the following can be determined: Vertex = (axis of symmetry, k). The Value of 'a' y = 1x2 y = 5x2 y = 0.2x2 y = -x2

--> a = 1 --> a > 1 --> 0 < a < 1 --> a < 0

--> basic parabola --> the opening gets more narrow --> the opening gets wider --> the parabola opens down

4.3 - General Quadratic Equation The vertex form of a parabola is: y = a(x - h)2 + k, where the vertex is (h, k). The axis of symmetry lies on the x-axis of the vertex. The maximum or minimum always occurs at the vertex. The 'h' value represents a horizontal translation, and the 'k' value represents a vertical translation. If the 'h' and 'k' values are 0, the vertex of the parabola is at the origin: (0, 0).

20

Sketching Technique Sketch the parabola y = -2(x + 3)2 + 4. 1. Identify the vertex: (-3, 4). 2. Before plotting the x and y axes, look at the vertex and the direction of the opening. Adjust the axes to compensate for the translations. 3. Plot the vertex. 4. Apply the scale factor ('a') the each y co-ordinate. (0, 0) is the vertex of the basic parabola, but the vertex of this parabola is (-3, 4). Go 1 unit to the left and 2 units down from the vertex. That is the second point. Since a parabola is symmetric, you can go 1 unit to the right and 2 units down from the vertex, creating the third point. x -2 -1 0 1 2

y 4 x 2 = -8 1 x 2 = -2 0x2=0 1x2=2 4x2=8

Co-ordinate (-2, -8) (-1, -2) (0, 0) (1, 2) (2, 8)

4.4 - Completing the Square Vertex form: Standard form:

y = 2(x - 3)2 + 4 y = 2(x2 - 6x + 9) + 4 y = 2x2 - 12x + 22

--> y = a(x - h)2 + k --> y = ax2 + bx + c

Advantages of Each Form Vertex Form vertex (h, k) axis of symmetry (x = h) direction of opening max/min value of y (y = k) sketching

Standard Form y-intercept (let x = 0) direction of opening

Therefore, there are more advantages of having the equation of a parabola in vertex form, rather than in standard form. Completing the Square 1. 2. 3. 4. 5. 6.

Group the terms containing the x variable, using square brackets: [ ]. Factor out the a-coefficient from the terms inside the [ ] brackets. Complete the square for the terms inside the [ ] brackets. Factor the perfect square trinomial into a perfect square binomial. Distribute the 'a' to remove the [ ] brackets. Simplify the constants.

21

Standard form:

Vertex form:

y = x2 - 10x + 3 y = [x2 - 10x] + 3 y = [x2 - 10x] + 3 y = [x2 - 10x + (-5)2 - (-5)2] + 3 y = [(x - 5)2 - (-5)2] + 3 y = (x - 5)2 - 25 + 3 y = (x - 5)2 - 22

--> Original equation. --> Step 1. --> Step 2. --> Step 3. --> Step 4. --> Step 5. --> Step 6.

4.4 - Word Problems Maximum Areas A rectangular pool is to be enclosed with 1 000m of fencing. What dimensions will produce a maximum area of the pool? Let w represent the width of the pool, and let l represent the length of the pool. Let A represent the area of the pool. 2w + 2l = 1,000 A = lw w = 500 - l

A = l(500 - l) A = - l2 + 500l A = - [l2 - 500l]

A = - [l2 - 500l + (-250)2 - (-250)2] A = - [(l - 250)2 - 62,500] A = (l - 250)2 + 62,500

The maximum area of the pool is 62,500m2. This maximum area occurs when the length of the pool occurs when the length is 250m, and the width is 250m. ************************************************************************ Numbers Find 2 numbers whose difference is 10 and whose product is a minimum. Let x represent the larger number. Let (x - 10) represent the smaller number. Let y represent the product. y = x(x - 10) y = x2 - 10x y = [x2 - 10x]

y = [x2 - 10x + (-5)2 - (-5)2] y = [(x - 5)2 - 25] y = (x - 5)2 - 25

The first number is -5, and the second number is 5. ************************************************************************

22

Profit and Revenue Revenue = Quantity x Selling Price Profit = Total Revenue - Total Cost The selling price of a Raptors ticket is \$40. The Air Canada Centre can seat 80,000 people when full. A survey indicates that if the price increased, attendance will drop by 10,000 people for every \$10 increase in the selling price. What ticket price will result in the greatest revenue? Let x represent the number of \$10 increments in the ticket price, and let y represent the revenue. y = (80,000 - 10,000x)(40 + 10x) y = 3,200,000 + 800,000x - 400,000x - 100,000x2 y = -100,000x2 + 400,000x + 3,200,000 y / 100,000 = -x2 + 4x + 32 y / 100,000 = - [x2 - 4x] + 32 y / 100,000 = - [x2 - 4x + (-2)2 - (-2)2] + 32 y / 100,000 = - (x - 2)2 + 36 y = -100,000(x - 2)2 + 3,600,000 A \$20 increase in the ticket price to \$60 will result in the maximum revenue of \$3,600,000.

5.1 - Quadratic Equations In chapter 4, quadratic functions took on the form of y = ax2 + bx + c. There were two variables (x and y), they were in the shape of a parabola, and had a maximum or a minimum. In chapter 5, the quadratic function is changed into a quadratic equation in the form 0 = ax2 + bx + c, by letting y = 0. There equations have only one variable (x). There are 3 ways of solving quadratic equations: 1. Graphing (y = ax2 + bx + c) 2. Factoring 3. Quadratic Formula As well, there are three different types of solutions: 1. 2 real and distinct solutions 2. 2 real and equal solutions 3. No real solutions

23

5.2 - Solving Quadratic Equations by Factoring The Zero Product Property states that if AB = 0, then either A = 0, B = 0, or A and B = 0. X-intercepts are also called roots. x2 + 2x - 24 = 0 (x + 6)(x - 4) (x + 6) = 0 || (x - 4) = 0

--> The quadratic equation. --> Factor into 2 binomials. --> Apply the Zero Product Property.

x = -6, x = 4

--> The two x-intercepts.

x2 - 25 = 0 (x + 5)(x - 5) (x + 5) = 0 || (x - 5) = 0 x = -5, x = 5

x2 - 13x = 0 x(x - 13) = 0 (x) = 0 || (x - 13) = 0 x = 0, x = 13

5.4 - The Quadratic Formula The quadratic formula is the general equation solved for x: ax2 + bx + c = 0 x2 + (b / a)x + c / a = 0 x2 + (b / a)x = -c / a x2 + (b / a)x + (b / 2a)2 = -c / a + (b / 2a)2 (x + b / 2a)2 = -c / a + (b / 2a)2 x + b / 2a = √[-c / a + (b / 2a)2] x = -b / 2a + √[-c / a + (b / 2a)2] x = {-b ± √[b2 - 4ac]} / 2a The Nature of the Roots The radicand of the quadratic formula (b2 - 4ac) is called the discriminant. To find the nature of the roots, look at the discriminant (D). This tells you if the parabola: • has 2 real and distinct roots --> D > 0 • 2 real and equal roots --> D = 0 • no real roots --> D < 0

5.5 - Complex Numbers The set of numbers beyond the real set is called the complex set of numbers: C. There are two parts to every complex number: a real part and an imaginary part. is the imaginary unit. = √-1. Therefore, 2 = -1. Now radicals with a negative radicand can be further simplified. √(-25) --> The square root of negative twenty-five. = √(-1) x √25 --> Can be written as the square root of -1 x the square root of 25. = 5 --> Can be simplified to 5.

24

2 / 3√6 = (2 / 3√6) x (√6 / √6) = 2√6 / 3 x 6 = √6 / 9

--> The original equation. --> Rationalize (1 = √6/√6). --> Simplify the fraction. --> Reduce the fraction.

(3 - 2) / (5 + ) = (3 - 2) / (5 + ) x (5 - ) / (5 - ) = (15 - 13 + 22)/ (25 - 2) = (13 - 13) / 26 = 13(1 - ) / 26 = (1 - ) / 2

--> The original equation. --> Multiply by a conjugate. --> Simplify. --> Replace the 2s by -1s. --> Factor 13 out of the numerator. --> Reduce the fraction.

6.1 - Introduction to Trigonometry Trigonometry is the study of the measurement of the sides and angles of triangles. Congruent triangles are triangles that have the same side lengths and angle measurements: they are identical in shape.  A

B

--> The corresponding angles are equal in degree measure: A = D, B E and C = F. C

--> The corresponding side lengths are equal: a = d, b = e, and c = f.

D

E

F

Classes of Angles • • • • •

Acute: < 90º Right: 90º Obtuse: > 90 and < 180º Straight: 180º Reflex: > 180º

Classes of Triangles • • • •

Scalene (3 different side lengths) Isosceles (2 same side lengths) Equilateral (all identical lengths) Right (one angle = 90º)

Pairs of Angles --> Complementary Angles: A + B = 90º A B

Supplementary Angles: C + D = 90º Opposing angles are equal in degree measure.

Corresponding Angles Theorem

--> When two parallel lines are intersected, the opposite angles are equal in degree measure. --> Look for the letter F.

Corresponding Angles Theorem

--> When two parallel lines are intersected, the acute angles are equal in degree measure. --> Look for the letter Z.

Co-interior Angles Theorem

--> When two parallel lines are intersected, the sum of two adjacent angles is 180º. --> Look for the letter C.

Ratios A ratio is a fraction: a / b, where a and b and integers, and b ≠ 0. A proportion is an equality of two ratios: a / b = c / d, or a : b = c : d. 3 : 4 = x : 16 --> The two original ratios. 4x = 48 --> Cross-multiply. x = 12 --> Solve for x.

26

Similar Triangles Similar triangles are triangles with identical corresponding angle measurements, as well as identical corresponding side length measurements. As well, if the ratio of side lengths of two similar triangles was x : y, then the ratio of their areas would be the square of the ratio of their side lengths: x2 : y2. To prove that two triangles are similar, you must prove that at least two of their corresponding angles are equal. This can be done using the aforementioned angle and line theorems.

6.2 - Other Angles

B

--> A is called the angle of elevation. It is the angle that a triangle makes with the horizontal, going up. --> ƹB is called the angle of depression. It is the angle that a triangle makes with the horizontal, going down. --> Both these angles are acute.

A

6.3 - The Tangent Ratio A

--> The three primary ratios are the Tangent ratio, the Sine ratio and the Cosine ratio. They can only be applied to right angle triangles. --> From the reference ƹA, a is the opposite side, c is the adjacent side, and b is the hypotenuse. Although the opposite and adjacent sides may change with the angle of reference, the hypotenuse always remains the same. It is the side opposite the right angle.

B

opposite side

C

The Tangent ratio can be expressed as: Tangent ƹA = Length of the side opposite ƹA / Length of the side adjacent ƹA or Tangent ƹA = opposite / adjacent

6.4 - The Sine Ratio Once the concept of a primary ratio has been understood, the remaining two rations are very simple and easy to grasp. The Sine ratio can be expressed as: Sine ƹA = Length of the side opposite ƹA / Length of the hypotenuse or Sine ƹA = opposite / hypotenuse

27

6.5 - The Cosine Ratio The Cosine ratio can be expressed as: Cosine ƹA = Length of the side adjacent ƹA / Length of the hypotenuse or Cosine ƹA = adjacent / hypotenuse A common mnemonic used to remember these three ratios is Soh Cah Toa:

Sine = opposite : hypotenuse Cosine = adjacent : hypotenuse

6.6 - Solving Right Triangles Solving a triangle means finding the measures of all unknown side lengths and angles. This can be done using the Pythagorean Theorem, primary ratios, or angle and line theorem.

B

11m

A

13m C a2 + b2 = c2 a2 + 121 = 169 a2 = 48 a≈7

--> The Pythagorean Theorem. --> Plug in the numbers. --> Simplify. --> Solve for a.

cosƹA = 11 : 13 ƹA ≈ 32º ƹC ≈ 58º

--> Use the Cosine ratio to find angle A. --> Solve for ƹA. --> Since the degree measure of all the angles of a triangle is 180º, you can solve for the remaining angle by solving 180 - 90 - 32.

ƹA ≈ 32º, ƹC ≈ 58º and a = 7m. --> Always include a concluding statement.

28