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
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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 —