MPM2D0 Final Exam

Linear Systems

A Linear Systemis composed of two or more linear relations.
To solve a Linear System you must determine which values of x and y are common to all lines. This occurs at the point of intersection.

Elimination

can only eliminate when the coefficient infront of both x's or both y's are equal and opposite signs

can only eliminate when the coefficient infront of both x's or both y's are equal and opposite signs

Substitution

Graphing

Only works when isolated for y for the y=mx+b format
m=rise/run
b=y intercept

Only works when isolated for y for the y=mx+b format
m=rise/run
b=y intercept

Application

1)Assign variables to each of the unknowns
2)Write 2 equations showing the relationships between the variables. Each equation should include both variables.
3)Solve the system of equations using any method (graphing,substitution,elimination)
4)Check your solution
5)Clearly communicate your final answer

Quadratic Functions/Polynomials

Terms

Variable: A number used to represent an unknown quantity that can vary

Coefficient: A number in which the variable multiplies with eg. coefficient = 5: (5x)

Term: A set of numbers that multiply with each other (eg. 5x^2)

Degree of polynomial: the exponent after the variable in a polynomial (eg. x^2y^3. degree=5 (2+3)

Degree of Term: The highest exponent in a term

1st/2nd Differences

If 1st relations jump by the same, it is linear
If 2nd relations jump by the same, it is expontial

ax²+bx+c

"Golden Formula": Everything derives from this

Roots

if (x,y) = (a,0) & (b,0), then y=(x-a)(x-b)
TO GET THE ROOTS:
1.Get rid of any fractions by multiplying each term by the LCD.
2.Write equation as ax2+ x + c = 0
3.Fully factor.
4.Set each factor equal to 0 and solve using the zero product propert

Parabolas

Properties of a Parabola

Properties of a Parabola

A parabola is a symmetrical graph, a 'U' shape. Its key components include:

Zeros: The points where the parabola intersects the x-axis (also called roots or x-intercepts).
Line of Symmetry: A vertical line that divides the parabola into two equal halves.
Minimum or Maximum: The lowest (minimum) or highest (maximum) point on the parabola.
Vertex: The point where the parabola meets the line of symmetry, which is also the minimum or maximum point.
Optimal Value: The y-coordinate of the vertex (either the minimum or maximum value).
Y-Intercept: The point where the parabola crosses the y-axis.

Forms of Quadratics, How to Factor

Forms of Quadratics, How to Factor

Standard Form: (ax²+bx+c)

Factoring Complex Trinomials

Factoring Complex Trinomials

To get from Standard Form to Factored Form:

Find two numbers that add to "b" and multiply to "c"

1: ex. x²+5x+6
b=5, c=6
numbers that add to 5 and multiply to 6:
3 and 2
Therefore x²+5x+6=(x+3)(x+2)

To get from Standard Form to Vertex Form:

1: if a isn't 1, factor out a so that the other steps work (ex, 2x²+8x+3=2(x²+4x)+3

2: Complete the square by:
Take half of the coefficient of x, square it, and add it inside the parentheses. Subtract the same value to balance the equation
ex: x²+4x, half of 4 is 2, and 2²=4, so
2(x²+4x+4-4)+3

3: Simplify the Equation by removing the negative in the end of the bracket
2(x²+4x+4)+3-8
=2(x²+4x+4)-5
*dont forget to multiply the negative by a (2)

4: Write in vertex form (in the bracket is a perfect square trinomial.
2(x²+4x+4)-5
=2(x+2)²-5

Factored Form: (x-r)(x-s)

Derrives from standard form (ax²+bx+c)

Use distributive property to get to standard form

Use distributive property to get to standard form

To get to Standard Form: Use Distributive Property, collect like-terms

To get to Vertex Form: Get to standard form and then (refer to "to get from Standard Form to Vertex Form)

Vertex Form: y=a(x-h)²+k

Vertex Form: y=a(x-h)²+k

Vertex (h,k)

h=-b/2a, which is the Axis Of Symmetry

k=f(h)=y

Vertex (h,k)=(-b/2a, substitute h for k)

a = +, opens up. a = -, opens down

Special Cases (Completing the Square)

The expanded expression is called a DIFFERENCE OF SQUARES. Both terms are perfect squares and the terms are separated by a mi

The expanded expression is called a DIFFERENCE OF SQUARES. Both terms are perfect squares and the terms are separated by a minus sign

To factor a difference of squares:
1)Two brackets that are the same but one with a plus (+) and one with a minus (-)
2)Square roots of perfect squares go in the brackets

In a perfect square trinomial, the first and last terms are perfect squares.
The middle term is twice the product of the square roots of the first and last terms

Quadratic Formula

Shows us the Roots of the Function

x=(-b±√(b²-4ac))/(2a)

THE DISCRIMINANT:
if √(b²-4ac)=0, 1 real solution
if √(b²-4ac)>0, 2 real solutions
if √(b²-4ac)<0, no real solutions

e

e

Distributive Property (Binomial Multiplication)

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

Transforming Quadratics

Translations

Vertical Translations:
y=x^2 + k: moves the vertex up (k>0)
y=x^2 - k: moves the vertex down (k<0)

Horizontal Translations:
y=(x+h)^2: moves the vertex left (h>0)
y=(x-h)^2: moves the vertex right (h<0)

Reflections and Stretches

Reflections:
if a in ax^2 is negative, the parabola opens up and reflects over x-axis (vice versa)

Vertical Stretch/Compression:
if a>0, the parabola is stretched
if 0<a<1 (a is in between 0 and 1), the parabola is vertically compressed

Step Pattern

"Step pattern" refering to the jump in y value for each one x value

A parabola with a scale factor of 1 has a "step pattern" of 1, 3, 5

If a=2, the jumps are 2, 6, 10

Trigonometry

Sine Law

Use when you have:
AAS
SSA
ASA

EXAMPLE: solving for side b:
a/sin35=42/sin45
a           = 42sin35/sin45
a           = 34.07
*same method with side c
*SATT

EXAMPLE: solving for side b:
a/sin35=42/sin45
a = 42sin35/sin45
a = 34.07
*same method with side c
*SATT for angle C

Cosine Law

When solving for cosine(x), use algebra

When solving for cosine(x), use algebra

Use when you have:
SSS
SAS

EXAMPLE: Solving for angle A:
4.8²  =8.5²+6.3² -2(8.5)(6.3)(cosA)
cosA=(-8.5²-6.3²+4.8²)/(-2(8.5)(6.3))
cosA=127/153
A      =

EXAMPLE: Solving for angle A:
4.8² =8.5²+6.3² -2(8.5)(6.3)(cosA)
cosA=(-8.5²-6.3²+4.8²)/(-2(8.5)(6.3))
cosA=127/153
A =33.9 degrees
*same method for other angles

SOH-CAH-TOA

Inverse Signs:
θ=sin-1(o/h)
θ=cos-1(a/h)
θ=tan-1(o/a)

Similarity/Congruence

If the triangles have same angles but different side lengths, they are similar:
AB/AC=EG/EF,   AC/BC=EF/GF,    AB/BC=EG/GF
∠A

If the triangles have same angles but different side lengths, they are similar:
AB/AC=EG/EF, AC/BC=EF/GF, AB/BC=EG/GF
∠A=∠E, ∠B=∠G, ∠C=∠F

If the traingles have same side lengths and angles, they are congruent:
AB=PQ,   AC=PR,   BC=QR
∠A=∠P,   ∠B=∠Q,    ∠C=∠R

If the traingles have same side lengths and angles, they are congruent:
AB=PQ, AC=PR, BC=QR
∠A=∠P, ∠B=∠Q, ∠C=∠R

Solving Similar Triangles

Solving Similar Triangles

Also: 
b^2=c^2-a^2
a^2=c^2-b^2

Also:
b^2=c^2-a^2
a^2=c^2-b^2

Analytic Geometry

Terms

Line Segment: A part of a line between two points

Midpoint: The point in the middle of a line segment

The line segment from a point in a triangle to the opposite side

Perpendicular: Two lines that form a 90 degree angle

Parallel: Two lines that never touch

Altitude: a line segment in a triangle from one vertex to the opposite side, creating a perpendicular line

Formulas

Slope of a line

Slope of a line

EXAMPLE: Points (3,5) & (-2,6)
slope=(6-5)/(-2-5)
slope=-1/7

Equation of a line: y=mx+b

Midpoint of a line segment

Midpoint of a line segment

EXAMPLE; Points (3,5) & (-7,-7)
midpoint=(-7+3)/2, (-7+5)/2
midpoint=(-2, -1)

√((x₂ – x₁)² + (y₂ – y₁)²)
Length of a Line Segment/Distance between Two Points

√((x₂ – x₁)² + (y₂ – y₁)²)
Length of a Line Segment/Distance between Two Points

EXAMPLE: Points (4,1) & -3,-5)
x₂ – x₁=4-(-3)=7
y₂ – y₁=1-(-5)=6
Length of Line=√(7² +6²)
=√83
=9.11

EXAMPLE: Centre at Origin, (x,y)=(2,2)
r^2=2^2+2^2
r^2=8
r=2.83

If Centre is at origin (0,0), then the equation is:
x^2+y^2=r^2

Tips

To Find The MEDIAN:
1)Find the midpoint of the opposite side
2)Find the slope of the line connecting the vertex to the midpoint of the opposite side
3)Calculatethe y-intercept of the line
4)Write the equation of the line.

To Find PERPENDICULAR LINE (RIGHT BISECTOR):
1: Flip the slope (eg. 2->1/2)
2: Change the sign (eg. 1/2->-1/2)
therefore: (2->-1/2)

To find the ALTITUDE:
1)Find the slope of the side opposite from the vertex
2)Find the slope of the altitude which is perpendicular to the side opposite from the vertex
3)Use the altitude’s slope and the point from the vertex to calculate the y-intercept of the altitude
4)Write the equation of the altitude

Classifying Figures

Classifying Figures

Exam Tips

Questions you don't understand, skip and come back to them

Dont rush, take your time

Think of questions fundamentally. If it doesn't make sense, it probably isn't right

Show work

Read questions carefully

Sleep well

Eat before

Don't forget Pencil, Eraser, Calculator, Ruler

Use "Therefore Statements" if it's a word problem

Line up equal signs

Dont stress out

Grade 9 Review

BEDMAS (Order of Operations: Brackets, Exponents, Division/Multiplication, Addition/Subtraction)

Y=MX+B (equation of a line
m=slope of line
b=y intercept
(x,y) can be anything on the x,y plane

Exponent Rules
a^b x a^c = a^(b+c)
a^b / a^c = a^(b-c)

Like Terms: Terms that can be grouped by adding or subtracting (x+y isn't xy.) (x + 3x=4x)