MPM 2D1 — Summary Sheet

Graphing Lines

Method IFind slope and y-interceptEquation must be y = mx + b formm is slope, b is y-intercept

Example I

= —y x — 4

slope= —%i’

rise = -2

y — intercept = (0-4) run = 3

start at (0-4) go down 2 right 3Graph is shown to the right

Method 2Find x and y interceptsEquation may be in any form

Example 23x

—

2y = —6

If the signs for the variable with thesame number are the same,subtract the two equations,otherwise add.(In this case we added)

3x — 2;’ = —6

—4

x-inty=03x = -6x = -2(-2,0)

datx0

= -6y=3(0,3)

Plot the two points and connectGraph is shown to the left

k;))FindIng the Point of Intersection

Technique 1: GraphingGraph each line using the techniques above. See where they cross. (In the above example, P01 is approximately (-3.2,- 1.8))

Technique 2: Substitution Technique 3: Elimination3x+2y=8 3x+2y=8

—lx+4y=9 —Jx+4y=9Use multiplication toget one variable

1 + 7 2 xl Th having the sameX 3x+2y=8 3x+2y=8 I numberinboth2;’ =—3x + 8 Isolate 1 variable in I equation

— lx + =x3

—3x + 12y = 23j> havey

= / x + 4 different sions

3x+23’=8

— lx + 4(%x + 4) = 9 — Sub into other equation — 3x + l2y = 27

—lx—6x+16=9

7x7

}_Solve

3’ + 2(54) = 8

; = —W +Sub answer into ohginal and 31 + 5 = 8

y54 J calculate 31=3

x=l

(1,54) The answer I(1,54) 4 The answer

Z , ‘9 :.

,a

.- Slope, Mldpplnt, Distance Fonnulas, Triangle Ce . ,

Slope= £!E= Y2 ‘i, Midpoint= (It +1 , 2), Distance= g(1, +(y, y1)2

run 2 2

Midsegment Line connecting two midpointsMedian: Connects a vertex of a triangle to the midpoint of the opposite sideAltitude: Connects a vertex of a triangle to the opposite side at a 90° anglePerpendicular Bisector: A line passing through the midpoint of a side of a triangle at a 90° angle

Shape Properties I ClassIfyIng Shapes

Parallelogram: Opposite sides are parallel.Rectangle: Opposite sides are parallel, has four go angles.Square: Opposite sides are parallel, has four go angles, all four sides have the same length.Rhombus Opposite sides are parallel, all four sides have the same length.All rectangles are also parallelograms, all squares are also rhombusesTo determine what type of shape you have, use the slope and distance formulas

: Properties of Quadratics (Paraboj-,. .

direction of opening: if a parabola opens up it has a U-shape, down it has aupside down U-shape.

vertex: the highest / lowest point on the parabolaaxis of symmetry: the line of symmetry on the parabola this vertical line

passes though the vertexzeros/roots: the point(s) where the parabola passes through the x

axis

—

In the graph to the right,

The direction of opening is UPThe vertex is (-1-4)The axis of symmetry is x = -1The zeros are (-3,0) and (1.0)

. Different Forms of the Parabola

FACTORED STANDARD VERTEXFORM FORM FORM

Expand Complete the Square

y=a(x—s)(x—t)4

y=ax2+bx+c y=a(x—h)2+k

Factor Expand

s and t are zeros a is the direction vertex is (h,k)a is direction of opening of opening a is direction of

opening

•::•:..; 2:4LtL Expanding and Factorin$

Expanding FactoringMultiply everything in one bracket by Technique 1: Common Factoring Technique 2: Simple Trinomialeverything in another.

+ 2 lx number in front of x2 is I

Example all terms are divisable by 3 and have an x x2 — óx + 8(—2x-i-7)(3x+4) GCFis3x —2+—4=—6=—6x2 —8x+21x+28 =3x(2x+7) —2x—4=8

= —6x2 +l3x+28 (x—2Xx—4)Technique 3: DecompositIon Technique 4: Duff of Squares

Example 2 number in front of x2 is not i both terms are perfect square

3(2x + 4)(—3x + 1) 2x2 ÷ Sx— 3 and are being subtracted

=3(—6x2+2x—12x+4) 6-i-—1=5 4x2+9

=3(—6x2—l0x+4) 6x—l=—6

=—18x2—30x+12 2x2+6x—lx—3 ‘J32x(x+3)—I(x+3) (2x+3)(2x—3)

(x+3)(2x— I)

Finding the Vertex

Method 1: Calculate the zeros, axis, vertex Method 2: Complete the Square

Example ExampleDetermine the vertex • use this information to graph Determine the vertex , use this information to graph

‘=2x2 —3x—20 y=2x2 + 12x+13Divide the numberinfrotofout

—8+ 5 =—3

y = 2? — 8x + 5x —20

Factor (if necessa) y = 2(x + 6x) + 134 I of first 2 tens

to get equation into b2

6 Determine the pertect square—8x 5 =—40

factored form=

= (3)2 =e number

y = 2x(x — 4) + 5(x — 4) 1 ii equation is alreadyy = 2(x2 + 6x + 9 — 9) + i

___

Add and subtract the numberDo this inside the brackets

‘= (2x + 5)(x Jfactored skip this step

I

_______________________

y = 2(x + 3)2— 5 +___—

Factor the first 3 terms

ZEROS Set each term equal Add up the leftoversto zero and solve to In this case 2 x -9 • 13 = -50=(2x+5)(x—4)

__________________________

get the zeros

}

Vertex is (3,5) 4 Remember to switch the xrdinale0=2xt5 OR O=x—4

____

OR x=42

To graph begin with y = x2 which has a vertex of (0,0) andAXIS OF SYMMETRY

adding the zeros12 )

Calculate the axis by Goes up over 1 unit (each way) up 1 over 1 up 3, over 1 up 5

and then dividing by Shift the vertex left 3 and down 5 to (-3-5)23

Stretch by 2 because the “a” value is positive 24

_______________________

Up 1, up 3, up 5 x2 Up 2, up 6, up 10VERTEX

Sub the axis into the Therefore,y=2(3<)2 —3<)— 20

y-coordinate of the

}

equauon to get the go over I unit (each way) and up 2, over 1 up 6, over I up 10

vertexThe graph is shown below. The graph of y =? is also

_____________________

shown

Vertex is169’

y=2(x+3)2 —5

. y=x2

.

.

. .

adjcos A = —

hjp

tan A =adj

These relationships can gy be used with a 90° angle.Sal—I CAH TOA can be used to help remember the ratios

Decide which angle and whici1 2 sides to useIn this case angle = 40, opp = 5, hyp = xopp and hyp is sinflip the fractions to gel x in the topcross multiply

sin,! sinB sinC a b cOR

a b c sinA srnfi sinC

Side a is opposite angle A, side B opposite angle b, etcTo use sin law, you must know one side-angle pair and you mustalso know one other side.

Examples Sin Law

COSINE LAW

a2 =j,2 ÷c2 —2(b)(c)(cos A)

b2 = a2 +c2 — 2(a)(c)(cos B)

= 2+ t2

— 2(a)(6)(cos C)

Used when you do nothave a side-angle pair

Examples Cosine Law

5cos A = —

7

A=cos(5/7)

A = 44.420

Decide which angle and which 2 sides to useIn this case angle = A, adj = 5, hyp = 7adj and hyp is cosuse cos’ to calculate an angle

a2 =5? 72 —2(5)(7)cos35

a225+49—70cos35

a2 =74—70cos35

a2 = 16.659

a = g16.659...a = 4.08

62=52 42 —2(4)(5)cosB

36 = 25 + 16— 4Ocos B

4ocos B = 5

cog B=

B=cos_h(5<o)

B = 82.820

The Quadratic Formula

If factoring does not work you can use this formula to Exampledetermine the zeroes or to determine if the zeroes exist. Determine the zeros of y = 3x2 — x — 5

, a=3,b=-1,c=-5The equation must be in the form y = ax + hx + c

- (- I) ± 1)2 - 4(3)(5)

2(3)2a I±gI+60

It is possible that there are no zeros/roots = 6

‘±JTIf b2 —4ac is negative there will be no zeros because you X

— 6cannot lake a square root of a negative number. 1±7.81 x = (1+7.81)16 OR x = (1—7.81)16

x=6 x=l.47 x=—114

. .47’ Trigonometry

Right Angle Trigonometry

sin A =hyp 0J

0aa0

Non-Right Angle TrigonomelrySIN LAW

hypotenuses

AAdjacent

Example 1: Calculating a Side

5

5sin40=—

x

sin40 5

x =5 x 1+ sin 40

x = 7.78

sin B — sin 3575

—117 sin 35B=sin I

“5

B = 52.42

x_8

sinj5 sinSO8 sin 35

1=Sin 50

=

Example 2: Calculating an Angle

7

MPM 2D1 UNIT 1: POLYNOMIALS REVIEW

PART A: Fill in the blank with the correct answer.

1. Simplify.

a) (_5x2y3_1onY5)=

b) (_2x2) =

— 3Oxy3zc) =

45xy5 z

2. Expand and simplify.

a) (x+5Xx—3)=

b) (2x+5X3x—4)=

c) (x+2)2 =

d) (3x—2v)2

3. Factor fully.

a) x2+óx—27 =

b) 9—4x2=

c) 25x2y — 5xy =

d) 3x2 +17x+1O =

e) x(m—2)± 4(2—rn) =

o x2+9

g)ti —14a2 +33 =

h) 20+x—x2=

i) 5x(a÷b)—(u+b)=

PART B: Show all your work

I. Expand and simplify.

a) 3x(2x + 5)2 b) (x — 3)ç2— 2x + 5) c) (3m + 2Xm —3)— (2,n

— lXm + 3)

d) 2(3x+l +3(2x—lX2x+1) e) (2x—7X3x—2)—5(x—4)2 +2(x2 +x—2)

I) 4(5x+2X5x_2)_x2 —(1—3x)2 +4xQ—2x)

2. Factor fully

a) ,z 15n2 +54 b) 20,1,2 —8i;i—12 c) 18y3 +60y2 +50y

d) ax+ay+3x+3y e) 2,1,2 —3p—6m+mp flI2x3y—4x2y2 +4y3

MPM 2D1 Unit 2: SYSTEMS OF EQUATIONS REVIEW

1. Solve the following systems.

x+4v=—3 2x+3y=14a) (Use Substitution.) b) (Use Elimination.)2x+5y=—3 8x—5y=—29

x 3’2(mn+I)—(n—4)=15

dc)3(mn—1)+4(n+2)=2 )

2(3x—1)—(y+4)=—7 x+y=1Oe)40 —2x)—3(3—y)=—12 O.3x+O.5y = 4.2

NIPNI 2D1 SYSTEMS OF EQUATIONS — WORD PROBLEMS Date:REVIEW

Without solving detennine whether Ihe following systems have one solution, no solutions or an) infinite number of solutions.

lOx—ôj’=—2 2x—3y=8a) — 1)= —I 4.v—oy = 10

2. SET UP the following word problems only by using let statements and/or charts. DO NOTSOLVE.

a) The difference of two numbers is 29. Four times the first plus twice the second is 64. Find thenumbers.

b) The sum of two numbers is 25. Five times the smaller is 3 more than double the larger. Find thenumbers.

c) A coin machine has $45.50 in dimes and quarters. There are 12 more dimes than quarters. Howmany of each are there?

d) The total of two investments is $50 000. Part is invested at 2.5%; the remainder at 4%. Theinterest on the investment was $425. How much was invested at each rate?

e) A chemist has an acid solution in two concentrations, 85% and 60%. He needs to make a 500 mlsolution which is 70% acid by volume. How much should he mix of each?

fl GORP is a mixture of trail mix and smarties. Trail mix sells for 59.00/kg and smarties for$14.50/kg. We would like 10kg of a mixture that costs $10.75/kg. How much of each should wemix?

MPM 2D1 UNIT 3 TRIGONOMETRY - REVIEW

1. Find the length of the indicated side using similar triangles. Round to the nearest tenth of aunit.

a)

B

D

b)

•1 -

M

750

x

30

E75

x

p

.800

F z

2. Find the length of the indicated side or the indicated angle in each of the following.

a) A x b)B D 10.4

F

16.3

c) d) ST

5.6

F

9.3

CE

8.9

8.7

E

x 2.2

R

b)AX7Z, ZX=73°,x=8, z=6.

e) A 1)

C

4 200

x

B cc 45D

D

3. a)SolveAPQRgivenp= 11 ,qloandr 14Find ZZ.

4. From a point 20 m from the base of a building, the angle of elevation to the top of the building is72°. Determine the height of the building.

5. From the top of a cliff the angle of depression to 2 boats is 25° and 40°. The boats are 125 mapart. Determine the height of the cliff

6. Determine the area of triangle DEF where e = 18cm, d = 18cm, and angle E = 65degrees.

7. Two buildings are 125 m apart. From the top of the taller building, the angles ofdepression to the top and bottom of the shorter building are 51 degrees and 36 degreesrespectively. Determine the height of

i) the taller buildingii) the shorter building.

MPM 2D1 UNIT 4: QUADRATIC FUNCTIONS - REVIEW

I. a) State whether the following are functions.b) State the domain and range.

I) {(—5—l), (—4,0), (—3,2), (—2,2)}

iii)

2. Complete the following table.

ii)

Equation ofM /M’

WhenFunction Vertex the Axis of Concavity Iuemn Max/Mm Domain Range

symmetry Occurs

y = + 4

y=—2(x—3)

y=—3(x+5)2 +2

V + 6 = + 2)2

b) Graph i) y+6=J(x+2)2

ii) i’=—2(x÷5) +4

:z:: 1rrrrrrE Z Z EZ ZJZIZJZEJZ

E4EEEJE2EEEJ

3. I) Graph the following equations. Show all steps. (Complete the Square)ii) Determine the x—intercepts.

a) y=2x2—16x±29

b) y=—3x2—12x—8

4. Determine the coordinates of the vertex for each of the following:

a) v=—2x1—8x—11 b) y=4x2±40x+98

5. Set tip the following problems only. Simplify the quadratic function.

a) The sum of 2 numbers is 30. Find the numbers if the sum of their squares is to be a minimum.

b) The difference of 2 numbers is 25. Find the numbers if their product is to be a minimum.

c) GO trains carry 5000 passengers per day from Milton to Toronto at $10.50 per ticket. A surveyshows that for every 50 cent increase in price there will be 100 less passengers. What ticket pricegives the maximum revenue?

d) A farmer has lOOm of fencing. wishes to fence off an area along a river as shown in the diagram.No fencing is required along the river. What dimensions give the maximum area?

RIVER

6. Iff(x)=x2 +óx+3 and g(x)=—Sx+5, find:

a) fF1) b) g(3)

7. Determine the equation of the following parabolas.

a) vertex (-5,1) and y-intercepts of 11

b) in the form y = a(x — 2)2 + k and passing through the points (3,5) and (0,9).

MPM 2D1 Unit 5: QUADRATIC EQUATIONS - REVIEW

I. Solve the following quadratic equations using the method of your choice.

a) lOx2 =20 b) (x—2)=12 c) x2+7x±6=O

d) 2x2—3x+l=O e) x2=4x 0 x2—2x+5=O

g) 2x2+3x=I h)x2+(x±I)2=(x+2)2

i) (x—lXx—2)+xG—O=x—l

j) x———=2 k)x+l X

2. Bill throws a baseball in the air. Its height, h metres, after t seconds is given by the formula:

h=l+20t—3r

a) If the ball was caught at the same height it was thrown, how long was it in the air?

b) If the ball was not caught, how long was it in the air?

MPM 2D1 QUADRATIC EQUATIONS - Word Problems

REVIEW

1. Set up the following problems only. Do not solve.

a) The sum of the squares oftwo consecutive odd integers is 290. Find the integers.

b) Find 2 integers whose sum is 96 and whose product is 1728.

c) A rectangular nuclear-waste facility is 150 m long and 80 m wide. A uniform strip, a safety zone,must be fenced around the facility. The area of the strip is twice the area of the facility.Determine the width of the strip?

it I •Unit 04NALYTIC GEOMETRY - REVIEW

Given the points P(—3,$) and Q(3,—2), determine:

I

j

a) the slope of the line segment PQ

c) the midpoint of the line 5egment PQ

b) the length of the line segment PQ

d) the equation of the line through P and Q

2. State the equation of the circle centered at (0,0) with the given radius.

i) 10

3. The vertices of a kite are G(2,5) , H(5,l), 1(8,2) and J(7,5).

ii) 5Ji

a) Graph the kite b) Veñ& that the diagonals of the kite areperpendicular.

MPM 2Dl Dale:

L

4, AABC has vertices A(2,l), B(8,3) and C(4,1). Determine:

a) the equation for SD, the median from S to AC

b) an equation for CE, the altitude from C to AD.

c) an equation forFG, the right bisector of BC.

5. The vertices of a triangle are A(—4,8), B(2,—4) and C(6,2). If M is the midpoint of AB and N is themidpoint of AC, verify that:

a) MN is parallel to BC

b) MN is half the length of BC

6. Find the shortest distance from the point (5,4) to the line x + v = 5. Express the distance as an exactvalue and as an approximate value to I decimal place.

i .:

. I —