MCR3U Grade 11 Functions Exam

MCR3U Exam Review Polynomials A polynomial is an algebraic expression with real coefficients and non-negative integer exponents.

A polynomial with 1 term is called a monomial, 7 . A polynomial with 2 terms is called a binomial, 3x2 − 9 . A polynomial with 3 terms is called a trinomial, 3x2 + 7− 9 .

The degree of the polynomial is determined by the value of the highest exponent of the variable in the polynomial.

e.g. 3x2 + 7− 9 , degree is 2.

For polynomials with one variable, if the degree is 0, then it is called a constant. If the degree is 1, then it is called linear. If the degree is 2, then it is called quadratic. If the degree is 3, then it is called cubic.

We can add and subtract polynomials by collecting like terms. e.g. Simplify.

(5x4 −x2 −2)−(x4 −2x3 +3x2 −5)

=5x4 −x2 −2−x4 +2x3 −3x2 +5

=5x4 −x4 +2x3 −x2 −3x2 −2+5

= 4x4 +2x3 −4x2 +3

To multiply polynomials, multiply each term in the first polynomial by each term in the second. e.g. Expand and simplify.

(x2 +4)(x2 −2x+3)

x4 −2x3 +3x2 +4x2 −8x+12

x4 −2x3 +7x2 −8x+12

Factoring Polynomials To expand means to write a product of polynomials as a sum or a difference of terms.

To factor means to write a sum or a difference of terms as a product of polynomials. Factoring is the inverse operation of expanding.

Expanding  (2x+3)(3x−7)=6x2 −5x−21

 Factoring

1

The negative in front of the brackets applies to every term inside the brackets. That is, you multiply each term by –1.

Sum or difference of terms

Product of polynomials

MCR3U Exam Review Types of factoring:

Common Factors: factors that are common among each term. e.g. Factor,

35m3n3 −21m2n2 +56m2n

=7m2n(5mn2 −3n+8)

Factor by grouping: group terms to help in the factoring process. e.g. Factor,

2

Each term is divisible by .

1+6x+9x2 is a perfect square trinomial

Group 4mx – 4nx and ny – my, factor each group

Recall – m

= –(m – n)

A:4mx+ny−4nxmy = 4mx−4nx+nymy = 4x(mn)+ y(nm) = 4x(mn)− y(mn) = (4− y)(n)

B:1+6x+9x2 −4y2 =(1+3x)2 −4y2

=[(1+3x)+2y][(1+3x)−2y] = (1+3x+2y)(1+3x−2y)

Difference of squares

Common factor

Factoring ax2 + bx Find the product of ac. Find two numbers that multiply to ac and add to b. e.g. Factor,

Product = 3(–6) = –18 = –9(2) Sum = – 7 = –9 + 2 Decompose middle term – 7xy into –9xy + 2xy.

Factor by grouping.

A:y2 +9y+14 = y2 +7y+2y+14

y(y+7)+2(y+7) = (y+2)(y+7)

B:3x2 −7xy−6y2 =3x2 −9xy+2xy−6y2

=3x(x−3y)+2y(x−3y) = (3x+2y)(x−3y)

Product = 14 = 2(7) Sum = 9 = 2 + 7

Sometimes polynomials can be factored using special patterns.

Perfect square trinomial

e.g. Factor, A:4p2 +12p+9

=(2p+3)2 Difference of squares

a2 + 2ab b2 = (b)(b) or a2 − 2ab b2 = (− b)(− b)

B:100x2 −80xy+16y2 =4(25x2 −20xy+4y2) = 4(5− 2y)(5− 2y)

a2 − b2 = (b)(− b) e.g.Factor, 9x2 −4y2 =(3x+2y)(3x−2y)

Things to think about when factoring:

• Is there a common factor? • Can I factor by grouping? • Are there any special patterns? • Check, can I factor x2 +bx+c? • Check, can I factor ax2 + bx ?

MCR3U Exam Review Rational Expressions

For polynomials and G, a rational expression is formed when GF ≠ 0 . e.g. 23x+7

3

21+14x+9

Simplifying Rational Expressions

e.g. Simplify and state the restrictions.

m2 −9 = (m+3)(m−3) m2 +6m+9 (m+3)(m+3) = (m+3)(m−3)

Factor the numerator and denominator. Note the restrictions.

Simplify.

(m+3)(m+3) = − 3 , ≠ −3

State the restrictions.

m+3 Multiplying and Dividing Rational Expressions

e.g. Simplify and state the restrictions.

A:x2 +7x× x2 +3x+2 x2 −1 x2 +14x+49

x(+ 7) (+1)(+ 2) = (x+1)(x−1)×(x+7)(x+7) x(x+7) (x+1)(x+2) = (x+1)(x−1)×(x+7)(x+7)

x(x+2) ,x≠±1,−7 (x−1)(x+7)

Adding and Subtracting Rational Expressions

e.g. Simplify and state the restrictions.

A:3+5 x2−4 x+2

= 3 + 5 (x−2)(x+2) x+2

= 3 + 5(− 2) (x−2)(x+2) (x+2)(x−2)

= 3+5x−10 (x+2)(x−2)

= 5− 7 , ≠ ±2 (x+2)(x−2)

Bx2 −9 ÷x2 −4x+3 x2 +5x+4 x2 +5x+4

= (+ 3)(− 3) ÷ (−1)(− 3) (+ 4)(+1) (+ 4)(+1) = (x+3)(x−3)×(x+4)(x+1)

(x+4)(x+1) (x−1)(x−3) =(x+3)(x−3)×(x+4)(x+1) (x+4)(x+1) (x−1)(x−3)

= (x+3), ≠ −4,±1,3 (x−1)

B: 2 − 3 x2−xy xyy2

= 2 − 3 x(− yy(− y)

= 2− 3x xy(xyxy(xy)

= 2y−3,x≠0,yy≠0 xy(− y)

Factor. Note restrictions.

Factor. Note restrictions.

Invert and multiply. Note any new restrictions.

Simplify.

State restrictions.

Simplify.

State restrictions.

Factor. Note restrictions. Simplify if possible.

Factor. Note restrictions. Simplify if possible.

Find LCD. Write all terms using LCD.

Find LCD. Write all terms using LCD.

Add.

Subtract. State restrictions.

State restrictions.

Note that after addition or subtraction it may be possible to factor the numerator and simplify the expression further. Always reduce the answer to lowest terms.

MCR3U Exam Review Radicals

e.g. , is called the radical sign, is the index of the radical, and is called the radicand. 3 is said to be a radical of order 2. 3 8 is a radical of order 3.

Like radicals: 5, 2 5, − 3 5 Unlike radicals: 5, 3 5, 3

Entire radicals: 8, 16, 29 Mixed radicals: 4 2, 2 3, 5 7

A radical in simplest form meets the following conditions:

4

Same order, like radicands

Different order

Different radicands

For a radical of order n, the radicand has no factor that is the nth power of an integer.

The radicand contains no fractions.

3 = 3 × 2 2 22

The radicand contains no factors with negative exponents.

The index of a radical must be as small as possible.

432= 32 =1×=3

a a Simplest form =a

aSimplest

8= 4×2 = 22×2 =22

Not 6

a−1=1 a

simplest form

=22 6

Simplest form

=22

=

6

2

Simplest form

=form To add or subtract radicals, you add or subtract the coefficients of each radical.

e.g. Simplify.

2 12−5 27+3 40=2 4×3−5 9×3+3 4×10 =2(2 3)−5(3 3)+3(2 10)

=4 3−15 3+6 10 =−11 3+6 10

Multiplying Radicals

e.g. Simplify.

( 2+2 3)( 2−3 3)=( 2)( 2)−( 2)(3 3)+(2 3)( 2)−(2 3)(3 3) =2−3 6+2 6−6(3)

=2−18−3 6+2 6 =−16− 6

Addition and Subtraction of Radicals

Express each radical in simplest form.

Collect like radicals. Add and subtract.

a× bab,a≥0,b≥0

Use the distributive property to expand

Multiply coefficients together. Multiply radicands together.

Collect like terms. Express in simplest form.

MCR3U Conjugates

Exam Review

) are called conjugates.

5

When conjugates are multiplied the result is a rational expression (no radicals). e.g. Find the product.

= 5−9(2) = 5−18 = −13

Dividing Radicals

Prime Factorization

Opposite signs

(a b c d )and (a b − Same terms Same terms

22 ( 5 + 3 2 )( 5 − 3 2 ) = ( 5 ) − (3 2 )

abba,a,bR,a≥0,b≥0

e.g. Simplify.

2 10+3 30=2 10+3 30 555

=2 10+3 30 55

=22+36

Factor a number into its prime factors using the tree diagram method.

Exponent Rules

e.g.

180 3 60

6 10 2325

Rule

Description

Example

Product

a×a=am+n

42 ×45 =47

Quotient

a÷a=amn

54 ÷52 =52

Power of a power

mn m×() =a

248 (3 ) = 3

Power of a quotient

aa  = ≠ 0

35 35  4  = 4 5

Zero as an exponent

a0 =1

70 =1

Negative exponents

a= 1 ,a≠0 am

9−2 = 1 92

Rational Exponents

m

an=a=(a)

nmn

4

343 273 = 27 =( 27)

4

MCR3U

e.g. Evaluate.

0 2 −2 −2 (3 + 3 ) = (1+ 9)

= 10−2 = 1

102 1

Exam Review

e.g. Simplify.

 b3 −2 b3(−2)

6

Follow the order of operations. Evaluate brackets first.

 2a−3   

− 2 = 2 = 2 + 2 = 4

= (2a−3 )−2 Power of a quotient. b−6

= 2−2 a−3(−2) Power of a product. = 22b−6

= 100 Solving Exponential Equations

e.g. Solve for x. 9x−2 − 8 = 73 Add 8 to both sides.

9x−2 = 73 + 8 Simplify.

a6 =4

a6b6

When the bases are the same, equate the exponents. Solve for x.

LS = 9x−2 − 8 = 94−2 − 8

= 81−8 = 73 = RS

RS = 73 = 4 checks

9x−2 =81 9x−2 =92

Functions

Note LS and RS are powers of 9, so rewrite them as powers using the same base.

Don’t forget to check your solution!

relation is a relationship between two sets. Relations can be described using: an equation an arrow diagram a graph

a table

y=3x2−7 g xy 8 -1 2 12

in words “output is three more than input”

a set of ordered pairs {(1,2),(0,3),(4,8)}

function notation f(x)=x2 −3x

0723 6-3 34

3 -5 -2

43

C: x−5 What value of will make – 5 = 0? = 5 The radicand cannot be less than zero, so Domain = {≥ 5, ∈ R}

The domain of a relation is the set of possible input values (values). The range is the set of possible output values (values).

e.g. State the domain and range.

A: {(1,2),(0,3),(4,8)}

Domain = {0, 1, 4} Range = {2, 3, 8}

B: 4 Looking at the graph we can see that does not go

2

below 0. Thus, Domain = Range =

Range = {≥ 0, ∈ R}

MCR3U Exam Review function is a special type of relation in which every element of the domain corresponds to exactly

one element of the range. x−7 and x2 +15 are examples of functions. = ± is not a function because for every

value of there are two values of y.

The vertical line test is used to determine if a graph of a relation is a function. If a vertical line can be passed along the entire length of the graph and it never touches more than one point at a time, then the relation is a function.

7

e.g. A: 4 This passes the vertical line

B: 4 2

The line passes through more than one point, so this relation fails the vertical line test. It is not a function.

2

Inverse Functions

test, so it is a function.

The inverse, −1 , of a relation, , maps each output of the original relation back onto the corresponding input value. The domain of the inverse is the range of the function, and the range of the inverse is the domain of the function. That is, if (ab) ∈ , then (ba) ∈ −1 . The graph of f−1 ( x)

is the reflection of the graph = e.g.Given f(x)=3x−1.

(x)

in the line x. Evaluate3f(2)+1

(2) +1 = 33(2) −1 +1  5 

=36−1+1  5 

5 = 3  5  + 1

= 3(1) + 1 3 (2) +1 = 4

Evaluate f(−3). (−3) = 3(−3) −1

5

f(−3)= −9−1 5

(−3) = −10 5

(−3) = −2

Determine −1(x). = 3−1

5

Replace all x’s with –3. Evaluate.

You want to find the value of the expression 3 (2) +1. You

are not solving for

(2) .

Evaluate 5 Rewritef(x)asy= 5 f−1(x)=

−1(2) 5+1

3 5(2) +1

3 10 +1

3−1

If you have not already determined −1(x) do so.

Using −1(x) , replace all x’s with 2. Evaluate.

= 3−1 5

5= 3−1 3y=5x+1 = 5+1

Interchange and y. −1 Solve for yf

(2) = = 3

−1(2) = 11 33

∴ −1 ( ) = 5 + 1 3

MCR3U Exam Review

e.g. Sketch the graph of the inverse of the given function (x). 444

yf−1(x) 222

8

Draw the line x.

Reflect the graph in the line x.

(x) -2 -4 -4 -4

-2

The inverse of a function is not necessarily going to be a function. If you would like the inverse to also be a function, you may have to restrict the domain or range of the original function. For the example above, the inverse will only be a function if we restrict the domain to {≥ 0, ∈ R} or {x≤ 0, ∈ R} .

Transformations of Functions

To graph af [k(− p)]+ from the graph (x) consider: – determines the vertical stretch. The graph (x) is stretched vertically by a factor of a. If < 0

then the graph is reflected in the x-axis, as well. – determines the horizontal stretch. The graph y(x) is stretched horizontally by a factor of 1. If

< 0 then the graph is also reflected in the y-axis. – determines the horizontal translation. If > 0 the graph shifts to the right by units. If < 0 then

the graph shifts left by units. – determines the vertical translation. If > 0 the graph shifts up byunits. If < 0 then the graph

shifts down by units. When applying transformations to a graph the stretches and reflections should be

4

2

-2

(x) -4

applied before any translations.

e.g. The graph of (x) is transformed into

= 3 (2− 4) . Describe the transformations.

First, factor inside the brackets to determine the values of and p.

= 3 (2(− 2)) = 3, = 2, = 2

There is a vertical stretch of 3.

A horizontal stretch of 12 . The graph will be shifted 2

units to the right.

e.g. Given the graph of (x) sketch the graph of = 2 (− (− 2)) +1

Reflect in y-axis. 22

Stretch vertically by a factor of 2.

4

4

-2

-4

-2

-4

4

2

Shift to the right by 2.

4

2

-2

Shift up by 1.

-2

This is the graph of = 2 (− (− 2)) +1

-4

-4

-2

4

Quadratic Functions

a>0

The graph of the quadratic function, (x) = ax2 + bx , is a parabola. When > 0 the parabola opens up. When < 0 the parabola opens down.

VertexForm: f(x)=a(xh)2 +k

The vertex is (h) . The maximum or minimum value is k.

2

-5 minimum 5

maximum

a<-0 4

The axis of symmetry is h. Factored Form: (x) = a(− p)(− qStandard Form: (x) =ax2 + bx c

The zeroes are and . The y-intercept is c.

Complete the square to change the standard form to vertex form. e.g.

(x) = −2x2 −12+ 7 Factor the coefficient of x2 form the terms with x2 and x.

f(x)=−2(x2 +6x)+7 f(x)=−2(x2 +6x+32 −32)+7 f(x)=−2(x2 +6x+32)−2(−32)+7 f(x)=−2(x+3)2 −2(−9)+7 f(x)=−2(x+3)2 +25 Simplify.

Maximum and Minimum Values

Divide the coefficient of by 2. Square this number. Add and subtract it. Bring the last term inside the bracket outside the brackets.

Factor the perfect square trinomial inside the brackets.

Vertex form, maximum/minimum value is k.

Factored form: e.g. Determine the maximum or minimum value of

(x) = (−1)(− 7) . The zeroes of (x) are equidistant from the axis of symmetry. The zeroes are x=1 and = 7 .

= 1 + 7 2

= 4

The axis of symmetry is = 4. The axis of symmetry passes through the vertex.

The x-coordinate of the vertex is 4. To find the y-coordinate of the vertex,evaluate f(4).

-2

(4) = (4 −1)(4 − 7) (4) = 3(−3) (4) = −9

The vertex is (4, − 9) . Because is positive ( = 1), the graph opens up.

The minimum value is –9.

Standard form: e.g. Determine the maximum or minimum value of (x) = −2x2 −10+10 without completing the square.

g(x) = −2x2 −10is a vertical translation of (x) = −2x2 −10+10 with y-intercept of 0.

g(x) = −2x(x+5)

= 0−5 = −2.5 2

= 0,−5 are the zeroes. = −2.5 is the x-coordinate of vertex.

Factor g(x) = −2x2 −10to determine zeroes, then find the axis of symmetry. Both

(x) and g(x) will have the same x

coordinates for the vertex. To find the y– coordinate for f(x) simply evaluate f(x) using the same x-coordinate.

(−2.5) = −2(−2.5)2 −10(−2.5) +10 (−2.5) = 22.5

The y-coordinate of vertex is 22.5. It is a maximum because the graph opens down.

Zeroes

To determine the number of zeroes of a quadratic function consider the form of the function. Vertex form: If and have opposite signs there are 2 zeroes (2 roots).

If and have the same sign there are no zeroes (0 roots). If = 0 there is one zero (1 root).

Factored form: Standard form:

(x) = a(− p)(− q→ 2 zeroes. The zeroes are and .

(x) = a(− p)2 → 1 zero. The zero is p.

Check discriminant. b2 − 4ac If < 0 there are no zeroes. If = 0 there is 1 zero. If > 0 there are 2 zeroes.

To determine the zeroes of from the standard form use the quadratic formula. For , ax2 + bx = 0 use = − ± b2 − 4ac to solve for x.

2a

Reciprocal functions

The reciprocal function of a function, , is defined as 1 . To help you graph = 1 , you should (x)

(x)

use the following: The vertical asymptotes of = 1 will occur where (x) = 0

(x) As (x) increases, 1 decreases. As (x) decreases, 1 increases.

(x) For (x) > 0, 1 > 0. For

1 < 0. (x)

(x) < 0, The graph of = 1 always passes through the points where (x) = 1 or (x) = −1.

(x)

(x) You may find it helpful to sketch the graph of (x) first, before you graph the reciprocal.

e.g. Sketch the graph of = 1 . x2 −4x

Look at the function (x) = x2 −4. Factorit. f(x)=x(x−4).

The zeroes are = 0, and = 4. The vertical asymptotes will be at = 0, and = 4.

You could sketch the graph of (x) = x2 − 4to see where the function increases and decreases, where x) = 1 or –1. Use the information above to help you sketch the reciprocal.

4

2

-2

-4

4

2

-2

-4

− 4x

5

5

y=1 x2 −4x

Vertical asymptotes

Exponential Functions

f(x) = 2x 6 4

2

In general, the exponential function is defined by the equation, aor f(x)=ax,a>0,xR.

Transformations apply to exponential functions the same way they do to all other functions.

f(x) = 2x-2+3 6 4

2

Exponential Growth and Decay

Population growth and radioactive decay can be modelled using exponential functions.

dt

N0 – initial amount

– time elapsed – doubling period

Decay: N(t) =  1 h

Growth:

) = 0 ( 2 )

N0 – initial amount

– time elapsed – half-life

N(t) – amount at time t A – future amount – present (initial) amount

– interest rate per conversion period – number of conversion periods

0  2 

N(t) – amount at time Calculating the future amount: P(1+ i)n

Calculating the present amount: A(1+ i)−Trigonometry

Compound Interest

Given a right angle triangle we can use the following ratios

Primary Trigonometric Ratios

sinθ = ry cosθ = rx tanθ = xy Reciprocal Trigonometric Ratios

cscθ=r= 1 secθ=r= 1 sinθ cosθ

Trigonometry of Oblique Triangles Sine Law

a=b=sinsinsinC

Can be used when you know ASA, AAS, SSA

Cosine Law

a2 =b2 +c2 −2bccosCan be used when you know SSS, SAS

r

x

A bc

y

cotθ=x= 1 tanθ

CaB

When you know SSA it is considered the ambiguous case.

C

Angle

< 90

> 90

Conditions

bsin A

bsin A

bsin A

# of Triangles

0

1

2C

A

ba b sinA

B

ab0a

B

a>b

1

b

A

Trigonometric Identities

Pythagorean Identity: sin2 θ +cos2 θ =1 e.g. Prove the identity. sin2 θ + 2cos2 θ −1 = cos2 θ

Quotient Identity: tanθ = sinθ cosθ

LS = sin2 θ + 2cos2 θ −1 Work with each side separately.

Since LS=RS then sin2 θ + 2cos2 θ −1 = cos2 θ is true for all values of θ .

= tanθ = sinθ

1

0.5

-0.5

-1

1

0.5

-0.5

-1

= sinθ

are shown below. = tanθ

= sin2 θ + cos2 θ + cos2 θ −1 = 1+ cos2 θ −1

Look for the quotient or Pythagorean identities. You may need to factor, simplify or split terms up. When you are done, write a concluding statement.

= cos2 θ = RS

Periodic Functions

A periodic function has a repeating pattern. The cycle is the smallest complete repeating pattern. The axis of the curve is a horizontal line that is midway between the maximum and minimum values of the graph. The equation is

= max value + min value . 2

Trigonometric Functions

The graphs of = sinθ , = cosθ , and

The period is the length of the cycle. The amplitude is the magnitude of the vertical distance from the axis of the curve to the maximum or minimum value. The equation is

= max value − min value 2

50

50

100 150

100 150

200 250

= cosθ

200 250

300 350

300 350

Period = 360 ̊ Amplitude = 1 Zeroes = 0 ̊,

180 ̊, 360 ̊…

= cosθ Period = 360 ̊

Amplitude = 1 Zeroes = 90 ̊, 270 ̊..5

-5

50

100 150 200 250 300 350

= tanθ Period = 180 ̊

Zeroes = 0 ̊, 180 ̊, 360 ̊… Vertical asymptotes = 90 ̊, 270 ̊..Transformations of Trigonometric Functions

Transformations apply to trig functions as they do to any other function. The graphs of sin (θ + b) + and cos (θ + b) + are transformations of the graphs

= sinθ and = cosθ respectively. The value of determines the vertical stretch, called theamplitude.

It also tells whether the curve is reflected in the θ -axis. The value of determines the horizontal stretch. The graph is stretched by a factor of 1. We can use this value to determine the period of the transformation of = sinθ or = cosθ .

The period of

= sin kθ

or

= cos kθ 360 = tan kθ 180 is > 0. The period of is > 0.

The value of determines the horizontal translation, known as the phase shift. The value of ddetermines the vertical translation. is the equation of the axis of the curve.

e.g. = cos2θ +1

e.g.

= 1sin(θ +45) 2

2 1.5 1 0.5

-0.5 -1

g(x) = cos(2⋅x)+1

1

0.5

-0.5

-1

50

100

150

200

250

300

350

400

f(x) = cos(x)

50 100 150

g(x) = 0.5⋅sin(x+45) 200 250 300 350 400

f(x) = sin(x)