MCR3U
1
Exam Review
Polynomials
A polynomial is an algebraic expression with real coefficients and non-negative integer exponents. A polynomial polynomial with 1 term is called a monomial, 7 x . A polynomial with 2 terms is called a binomial, 3 x 2 − 9 . A polynomial with 3 terms is called a trinomial, 3 x 2 + 7 x − 9 . he degree of the polynomial is determined by the val!e of the highest exponent of the variable in the polynomial. e.g. 3 x 2 + 7 x − 9 , degree is 2. "or polynomials with one variable, if the degree is #, then it is called a constant. co nstant. $f the degree is 1, then it is called linear. $f the degree is 2, then it is called %!adratic. $f the degree is 3, then it is called c!bic. &e can add and s!btract polynomials po lynomials by collecting li'e terms. e.g. (implify. (implify.
( * x − x − 2) − ( x − 2 x + 3 x − *) )
2
)
3
2
= * x ) − x 2 − 2 − x ) + 2 x 3 − 3 x 2 + * = * x ) − x ) + 2 x 3 − x 2 − 3 x 2 − 2 + * = ) x ) + 2 x 3 − ) x 2 + 3
he negative in front of the brac'ets applies to every term inside the brac'ets. hat is, yo! m!ltiply each term by 1.
o m!ltiply polynomials, m!ltiply each term in the first polynomial by each term in the second. e.g. +xpand and simplify.
( x + ))( x − 2 x + 3) 2
2
= x ) − 2 x3 + 3 x 2 + ) x 2 − x + 12 = x ) − 2 x3 + 7 x 2 − x + 12 Factoring Polynomials
o expand means to write a prod!ct of polynomials as a s!m or a difference of terms. o factor means to write a s!m or a difference of terms as a prod!ct of polynomials. "actoring is the inverse operation of expanding.
+xpanding
( 2 x + 3) ( 3 x − 7) = - x 2 − * x − 21 "actoring
/rod!ct of polynomials
(!m or difference of terms
MCR3U
2
Exam Review
Types of factoring: Common Factors: factors that are common among each term. e.g. "actor,
− 21m 2 n 2 + *- m 2 n = 7 m 2 n (*mn 2 − 3n + )
3*m 3n 3
+ach term is divisible by .
Factor by groping: gro!p terms to help in the factoring process. e.g. "actor, ;ro!p )mx )nx and A 0 )mx + ny − )nx − my
= )mx − )nx + ny − my = ) x( m − n ) + y ( n − m ) = ) x( m − n ) − y ( m − n ) = ( ) x − y )( m − n )
B 0 1 + - x + 9 x
18 x89 x2 is a perfect s%!are trinomial
− ) y 2 = 1 + 3 x 2 − ) y 2 ifference of s%!ares = 41 + 3 x + 2 y41 + 3 x − 2 y = 1 + 3 x + 2 y1 + 3 x − 2 y
ny my, factor each gro!p :ecall n m m n
2
6ommon factor
Factoring ax 2 + bx + c "ind the prod!ct of ac. "ind two n!mbers that m!ltiply to ac and add to b. e.g. "actor,
+ 9 y + 1) = y 2 + 7 y + 2 y + 1) = y y + 7 + 2 y + 7 = y + 2 y + 7
A 0 y 2
/rod!ct 1) 27 (!m 9 2 8 7
− 7 xy − - y 2 = 3 x 2 − 9 xy + 2 xy − - y 2 = 3 x x − 3 y + 2 y x − 3 y = 3 x + 2 y x − 3 y
B 0 3 x
2
/rod!ct 3 1 92 (!m 7 9 8 2 ecompose middle term 7 xy into 9 xy 8 2 xy. "actor by gro!ping.
(ometimes polynomials can be factored !sing special patterns . Perfect s!are trinomial e.g. "actor,
+ 12 p + 9 = 2 p + 3 2
A 0 ) p 2
"ifference of s!ares 9 x 2 − ) y 2 e.g. "actor,
a2
+ 2ab + b 2 = a + ba + b − # xy + 1- y 2 = ) 2* x 2 − 2# xy + ) y 2 = )* x − 2 y * x − 2 y
B 0 1## x 2
− b 2 = a + ba − b = 3 x + 2 y 3 x − 2 y a2
T#ings to t#in$ abot w#en factoring: • $s there a common factor5 • 6an $ factor by gro!ping5 • Are there any special patterns5 • 6hec', can $ factor x 2 + bx + c 5 • 6hec', can $ factor ax 2 + bx + c 5
or a 2 − 2ab + b 2 = a − b a − b
MCR3U
3
Exam Review
Rational Expressions
"or polynomials F and G, a rational expression is formed when
F G
,G
≠ #.
3 x + 7
e.g.
21 x
2
+ 1) x + 9
%implifying Rational Expressions e.g. (implify and state the restrictions.
− 9 = m + 3m − 3 + -m + 9 m + 3m + 3 = m + 3m − 3 m + 3m + 3 m−3 , m ≠ −3 = m+3
m2 m
2
"actor the n!merator and denominator.
(implify. (tate the restrictions.
Mltiplying and "ividing Rational Expressions e.g. (implify and state the restrictions.
+ 7 x x + 3 x + 2 × 2 x − 1 x + 1) x + )9 x x + 7 x + 1 x + 2 "actor. = ×
A 0
x
2
2
2
2 9 x − ) x + 3 − B 0 2 ÷ x + * x + ) x 2 + * x + ) x + 3 x − 3 x − 1 x − 3 "actor. = ÷ x + ) x + 1 x + ) x + 1
x
2
&dding and %btracting Rational Expressions e.g. (implify and state the restrictions. A 0
3 x
= = = =
2
−)
+
*
x + 2 3
x − 2 x + 2 3 x − 2 x + 2
+ +
*
"actor.
x + 2 * x − 2 x + 2 x − 2
3 + * x − 1# x + 2 x − 2 * x − 7 , x x + 2 x − 2
B 0
2 x
= "ind =6. &rite all terms !sing =6.
= =
Add.
≠ ±2
2
− xy
−
"actor.
3 xy − y
2 x x − y
−
2 y xy x − y 2 y − 3 x xy x − y
2
3
"ind =6. &rite all terms !sing =6.
y x − y
−
3 x xy x − y
, x
≠ #, y,
y
≠#
(!btract. (tate restrictions.
(tate restrictions.
MCR3U
)
Exam Review
Radicals
e.g.
n
a,
is called the radical sign, n is the index of the radical, and a is called the radicand. 3 is said to be a radical of order 2. 3 is a radical of order 3.
=i'e radicals0
*, 2
*,
−3
>nli'e radicals0
*
*,
3
*,
3
(ame order, li'e radicands
+ntire radicals0 ?ixed radicals0
, )
1- ,
2, 2
ifferent order
29
3, *
ifferent radicands
7
A radical in simplest form meets the following conditions0 "or a radical of order n, the radicand has no factor that is the nth power of an integer. = )× 2
he radicand contains no fractions. 3 3 2
2
=
2 2
2
=
1
=
a
-
= =
2
-
=
(implest form
×
he radicand contains no factors with negative exponents. 1 − a 1 = a
22 (implest form
-
=
2
a a
×
he index of a radical m!st be as small as possible. )
32
= =
32 3
a (implest form
a
2
a
(implest form
a
&ddition and %btraction of Radicals o add or s!btract radicals, yo! add or s!btract the coefficients of each radical.
e.g. (implify. 2 12 − * 27 + 3
)#
= 2 )× 3 − * 9× 3 + 3 = 2( 2 3 ) − *( 3 3) + 3( 2 = ) 3 − 1* 3 + - 1# = −11 3 + - 1#
) × 1# 1#
+xpress each radical in simplest form.
) 6ollect li'e radicals. Add and s!btract.
Mltiplying Radicals a×
b
=
ab , a
e.g. (implify. ( 2 + 2 3 )( 2 − 3
≥ #, b ≥ #
3
)=(
2
)(
2
) −(
)(
) + ( 2 3)(
2 3 3
= 2 − 3 - + 2 - − -( 3) = 2 − 1 − 3 - + 2 = −1- − -
) −( 2 3)( 3 3)
2
>se the distrib!tive property to expand
?!ltiply coefficients together. ?!ltiply radicands together.
6ollect li'e terms. +xpress in simplest form.
MCR3U
*
Exam Review
Con'gates
(
Cpposite signs a b + c d and a b
(
(ame terms
−c
d are called con@!gates.
(ame terms
&hen con@!gates are m!ltiplied the res!lt is a rational expression no radicals. e.g. "ind the prod!ct.
(
*
+3
2 )( *
−3
2)
= ( * ) − (3 = * − 9 2 = * − 1 = −13 2
2)
2
"ividing Radicals a b
=
a b
, a , b∈ R , a
e.g. (implify.
≥ #, b ≥ #
+3
2 1#
3#
=
*
2 1# *
=2
1#
=2
2
*
+
3 3# * 3#
+3
+3
* -
Prime Factori(ation
e.g.
1#
"actor a n!mber into its prime factors !sing the tree diagram method.
3
# 2
1# 3
2
*
Exponent Rles Rle /rod!ct !otient /ower of a power
"escription
/ower of a %!otient
a = a n , b ≠ # b b n
Bero as an exponent
a#
:ational +xponents
e.g. +val!ate.
Example 2 * 7 ) ×) = )
m+ n
×a = a a ÷ a n = a m −n ( a m ) n = a m× n a
m
n
÷ *2 = *2 ( 32 ) ) = 3
m
*)
n
=1
a −m
= n
=1
7#
1 a
m
,a ≠ #
m n
*
3 = 3* ) )*
m
a = a =
( a) n
9 m
e.g. (implify.
−2
=
1 9
2
3
) 27 =
)
27 3 =
(
3
27
)
)
MCR3U
Exam Review −2
−2
( 3 + 3 ) = (1 + 9 ) − #
2
2
= 1#−2 = =
1
b3 b 3 −2 −3 = −3 −2 2a 2a
"ollow the order of operations. +val!ate brac'ets first.
=
2
1# 1
=
1##
=
/ower of a %!otient.
b −2 −2 a −3 −2
/ower of a prod!ct.
2 2 b −a) a -b -
%olving Exponential E!ations e.g. (olve for x. 9 x−2
− = 73 Add to both sides. x− 2 9 = 73 + (implify. x− 2 9 = 1
x − 2
=2 x = 2 + 2 x = )
!sing the same base.
&hen the bases are the same, e%!ate the exponents. (olve for x.
LS = 9 x −2
− = 9 )− 2 − = 1 − = 73 = RS
RS = 73
x
= ) chec's
onIt forget to chec' yo!r sol!tionJ
Fnctions
A relation is a relationship between two sets. :elations can be described !sing0 an e%!ation an arrow diagram a graph 2 g y = 3 x − 7 # 3
in words Do!tp!t is three more than inp!tE a set of ordered pairs
-1 7 -3 -*
a table x y 1 2 2 3 3 ) ) 3
2
-2
G1, 2, #, 3, ), F
f!nction notation f x = x 2
− 3 x
he domain of a relation is the set of possible inp!t val!es x val!es. he range is the set of possible o!tp!t val!es y val!es. e.g. (tate the domain and range. A0 G1, 2, #, 3, ), F H0 omain G#, 1, )F :ange G2, 3, F
2
=oo'ing at the graph we can see that y does not go below #. h!s, omain R :ange
60
y
=
x−*
&hat val!e of x will ma'e x * #5 x * he radicand cannot be less than Kero, so omain G x L x ≥ *, x ∈ R F :ange G y L y ≥ #, y ∈ R F
MCR3U
7
Exam Review
A fnction is a special type of relation in which every element of the domain corresponds to exactly one element of the range. y = x − 7 and y = x 2 + 1* are examples of f!nctions. y = ± x is not a f!nction beca!se for every val!e of x there are two val!es of y. he vertical line test is !sed to determine if a graph of a relation is a f!nction. $f a vertical line can be passed along the entire length of the graph and it never to!ches more than one point at a time, then the relation is a f!nction. e.g. A0 H0 he line passes thro!gh more his passes the than one point, so this relation vertical line fails the vertical line test. $t is test, so it is a not a f!nction. f!nction. 2
2
)nverse Fnctions he inverse, f −1 , of a relation, f , maps each o!tp!t of the original relation bac' onto the corresponding inp!t val!e. he domain of the inverse is the range of the f!nction, and the range of the inverse is the domain of the f!nction. hat is, if a, b ∈ f , then b, a ∈ f −1 . he graph of − y = f 1 x is the reflection of the graph y = f x in the line y = x .
e.g. ;iven f x
3x − 1
=
*
+val!ate f −3 . f −3
=
f −3
=
f −3
=
f −3
=
. +val!ate 3 f 2 + 1
3 −3 − 1 :eplace all * with 3. − 9 − 1 xIs +val!ate. * − 1# * −2
etermine f −1 x . y x
3 x − 1
= =
y
=
3x
* y. 3 y − 1
−1
*
∴ f
x
=
f −1 2
= =
f −1 2
3
−1
f −1 x
$nterchange x and
= 3 y = * x + 1 * x + 1 y = * x
=
*x + 1
=
Mo! want to find the val!e of the expression 3 f 2 + 1 . Mo! are not solving for f 2 .
+val!ate f −1 2
:ewrite f x as
* 3 y − 1
32 − 1 + 1 * − = 3 - 1 + 1 * * = 3 + 1 * = 31 + 1 3 f 2 + 1 = ) 3 f 2 + 1 = 3
* x + 1 3 * 2 + 1 3 1# + 1
$f yo! have not already determined − f 1 x do so. −1
>sing f x , replace all xIs with 2.
3 11 3
3
4
−1
y = f x
e.g. ('etch the graph of the inverse of the given f!nction y = f x . 2
2
:eflect the graph in the line y x.
raw the line y x.
y = f x
-2
2
-2
-2
MCR3U
Exam Review
he inverse of a f!nction is not necessarily going to be a f!nction. $f yo! wo!ld li'e the inverse to also be a f!nction, yo! may have to restrict the domain or range of the original f!nction. "or the example above, the inverse will only be a f!nction if we restrict the domain to G x L x ≥ #, x ∈ R F or G x L x ≤ #, x ∈ R F . Transformations of Fnctions
o graph y = af 4k x − p + q from the graph y = f x consider0 a determines the vertical stretch. he graph y = f x is stretched vertically by a factor of a. $f a N # then the graph is reflected in the x-axis, as well. k determines the horiKontal stretch. he graph y = f x is stretched horiKontally by a factor of
1 . k
$f k N # then the graph is also reflected in the y-axis. p determines the horiKontal translation. $f p O # the graph shifts to the right by p !nits. $f p N # then the graph shifts left by p !nits. q determines the vertical translation. $f q O # the graph shifts !p by q !nits. $f q N # then the graph shifts down by q !nits. 4
&hen applying transformations to a graph the stretches and reflections sho!ld be applied before any translations. e.g. he graph of y = f x is transformed into y = 3 f 2 x − ) . escribe the transformations.
-
-4
4 4
(hift to the right by 2.
"irst, factor inside the brac'ets to determine the val!es of k and p.
= 3 f ( 2( x − 2) ) a = 3, k = 2, p = 2
(hift !p by 1. 2
y
here is a vertical stretch of 3. A horiKontal stretch of
-2
-2
-
-4
1 . 2
he graph will be shifted 2 !nits to the right. e.g. ;iven the graph of y = f x s'etch the graph of y
= 2 f ( − ( x − 2) ) + 1
y = f x
4
2
(tretch vertically by a factor of 2. -
:eflect in y-axis.
-2
his is the graph of
= 2 (− ( x − 2 +
4
aO#
*adratic Fnctions
2
he graph of the %!adratic f!nction, f x = ax + bx + c , is a parabola. &hen a > # the parabola opens !p. &hen a < # the parabola opens down.
2
minim!m
-5
5
maxim!m -2
aN#
2
+ertex Form: f x = a x − h + k he vertex is h, k . he maxim!m or minim!m val!e is k . he axis of symmetry is y h. 2 Factored Form: f x = a x − p x − q %tandard Form: f x = ax + bx + c he Keroes are x = p and x = q . he y-intercept is c.
-4
Complete t#e s!are to change the standard form to vertex form. e.g.
= −2 x 2 − 12 x + 7 "actor the coefficient of x2 form the terms with x2 and x. 2 f x = −2( x + - x ) + 7 ivide the coefficient of x by 2. (%!are this n!mber. Add and s!btract it. 2 2 2 f x = −2( x + - x + 3 − 3 ) + 7 f x = −2( x 2 + - x + 32 ) − 2−32 + 7 Hring the last term inside the brac'et o!tside the brac'ets. 2 f x = −2 x + 3 − 2−9 + 7 "actor the perfect s%!are trinomial inside the brac'ets. f x = −2 x + 3 2 + 2* f x
(implify.
Maximm and Minimm +ales
Pertex form, maxim!mQminim!m val!e is k . "actored form0 e.g. etermine the maxim!m or minim!m val!e of f x = x − 1 x − 7 . he Keroes of f x are e%!idistant from the axis of symmetry. he Keroes are x = 1 and x = 7 1+ 7 . x = 2 he axis of symmetry is x ). he axis of symmetry passes thro!gh the vertex. x = ) he x-coordinate of the vertex is ). o find the y-coordinate of the vertex, eval!ate f ) .
= ) − 1) − 7 f ) = 3−3 he vertex is !p. f ) = −9 f )
), − 9 . Heca!se a is positive a
= 1 , the graph opens
he minim!m val!e is 9.
(tandard form0 e.g. etermine the maxim!m or minim!m val!e of f x = −2 x 2 − 1# x + 1# witho!t completing the s%!are. g x = −2 x 2 − 1# x is a vertical translation of f x = −2 x 2 − 1# x + 1# with y-intercept of #. x = #, − * are the Keroes. g x = −2 x x + * 2 "actor g x = −2 x − 1# x to determine #−* = = −2.* x = −2.* is the x-coordinate of 2 f −2.* = −2 −2.*2 − 1#−2.* + 1# vertex. f −2.* = 22.* x
Keroes, then find the axis of symmetry. Hoth f x and g x will have the same xcoordinates for the vertex. o find the ycoordinate for f(x) simply eval!ate f(x) !sing the same x-coordinate.
he y-coordinate of vertex is 22.*. $t is a maxim!m beca!se the graph opens down. ,eroes
o determine the n!mber of Keroes of a %!adratic f!nction consider the form of the f!nction. Pertex form0 $f a and k have opposite signs there are 2 Keroes 2 roots. $f a and k have the same sign there are no Keroes # roots. $f k # there is one Kero 1 root. "actored form0 f x = a x − p x − q - 2 Keroes. he Keroes are x = p and x = q . f x = a x − p 2 - 1 Kero. he Kero is x p. (tandard form0 6hec' discriminant. D = b 2 − )ac $f D < # there are no Keroes. $f D = # there is 1 Kero. $f D > # there are 2 Keroes. o determine the Keroes of from the standard form !se the !adratic formla . "or , ax
2
+ bx + c = # !se x =
−b±
b2 2a
− )ac to solve for x.
Reciprocal fnctions
he reciprocal f!nction of a f!nction, f , is defined as
1 1 . o help yo! graph y = , yo! sho!ld f f x
!se the following0 1
he vertical asymptotes of y =
f x
will occ!r where f x = #
1 1 decreases. As f x decreases, increases. f x f x 1 > # . "or f x < # , 1 < # . "or f x > # , f x f x 1 he graph of y = always passes thro!gh the points where f x = 1 or f x = −1 . f x Mo! may find it helpf!l to s'etch the graph of y = f x first, before yo! graph the reciprocal.
As f x increases,
e.g. ('etch the graph of y
=
1 x
2
− ) x
y
.
=oo' at the f!nction f x = x − ) x . "actor it. f x = x x − ) . he Keroes are x #, and x ). he vertical asymptotes will be at x #, and x ). 2 Mo! co!ld s'etch the graph of f x = x − ) x to see where the f!nction increases and decreases, where f x = 1 or 1. >se the information above to help yo! s'etch the reciprocal.
=
x
2
−
) x
2
2
5
-2
-
2
y 5
-2
f(x) = 2x
6
Exponential Fnctions
2
1
=
x
2
−
f(x) = 2x-2+3
$n general, the exponential f!nction y = a x asymptotes is defined by the e%!ation, Pertical x or f x = a , a > #, x ∈ R . ransformations apply to exponential f!nctions the same way they do to all other f!nctions.
) x 6
-
2
Exponential .rowt# and "ecay /op!lation growth and radioactive decay can be modelled !sing exponential f!nctions. t
;rowth0 N t = N 2 d #
t
1 h ecay0 N t = N # 2
N # - initial amo!nt
N # - initial amo!nt
t time elapsed d do!bling period N t - amo!nt at time t
t time elapsed h half-life N t - amo!nt at time
t Compond )nterest 6alc!lating the f!t!re amo!nt0 A = P 1 + i n amo!nt 6alc!lating the present amo!nt0 P = A1 + i − n
A f!t!re amo!nt
P present initial
i interest rate per conversion period n n!mber of conversion periods
Trigonometry
;iven a right angle triangle we can !se the following ratios Primary Trigonometric Ratios sin θ
y
=
cos θ
r
=
x
tan θ
r
r y
= y x
Reciprocal Trigonometric Ratios cscθ
=
r
=
y
1
secθ =
sin θ
r x
=
1 cos θ
cot θ
=
x y
=
x
1 tan θ
A
Trigonometry of /bli!e Triangles %ine 0aw a sin A
=
b sin B
=
c
b
sin
c
6an be !sed when yo! 'now A(A, AA(, ((A Cosine 0aw a
2
= b2 + c 2 − 2bc cos A
6
6an be !sed when yo! 'now (((, (A(
&hen yo! 'now ((A it is considered the ambig!o!s case. Angle ∠ A < 9#
6onditions
a < b sin A a = b sin A a > b sin A
R of riangles # 1 2
a
C
b
a b sinA
A
B
H
C
∠ A > 9#
a ≤ b a > b
# 1
a b A
B
Trigonometric )dentities
!otient $dentity0 tan θ =
/ythagorean $dentity0 sin 2 θ + cos2 θ = 1 e.g. /rove the identity. sin LS = sin
2
θ
+ 2 cos
2
+ 2 cos θ − 1 = sin θ + cos 2 θ + cos 2 θ − 1 = 1 + cos 2 θ − 1 = cos 2 θ = RS 2
2
θ
θ
− 1 = cos
2
sin θ cos θ
θ
&or' with each side separately. =oo' for the %!otient or /ythagorean identities. Mo! may need to factor, simplify or split terms !p. &hen yo! are done, write a concl!ding statement.
2
(ince =(:( then sin 2 θ + 2 cos2 θ − 1 = cos2 θ is tr!e for all val!es of
Periodic Fnctions A periodic f!nction has a repeating pattern. he cycle is the smallest complete repeating pattern. he axis of t#e crve is a horiKontal line that is midway between the maxim!m and minim!m val!es of the graph. he e%!ation is
=
y
θ .
he period is the length of the cycle. he amplitde is the magnit!de of the vertical distance from the axis of the c!rve to the maxim!m or minim!m val!e. he e%!ation is a
=
max val!e − min val!e 2
max val!e + min val!e . 2
Trigonometric Fnctions he graphs of y = sin θ , y = cos θ , and y = tan θ are shown below. y
1
= sin θ
y
0.5
50
100
150
200
250
300
350
-0.5
= tan θ
y = sin θ /eriod 3#S Amplit!de 1 Beroes #S, 1#S, 3#ST
-1
50
100
150
200
25
30
-5
y
1
= cos θ
0.5
50
-0.5
5
100
150
200
250
300
350
y = cos θ /eriod 3#S Amplit!de 1 Beroes 9#S, 27#ST
y = tan θ /eriod 1#S Beroes #S, 1#S, 3#ST Pertical asymptotes 9#S, 27#ST
-1
Transformations of Trigonometric Fnctions
ransformations apply to trig f!nctions as they do to any other f!nction. he graphs of y = a sin k θ + b + d and y = a cos k θ + b + d are transformations of the graphs y = sin θ and y = cos θ respectively. he val!e of a determines the vertical stretch, called the amplitde . $t also tells whether the c!rve is reflected in the θ -axis.
350
he val!e of k determines the horiKontal stretch. he graph is stretched by a factor of this val!e to determine the period of the transformation of y = sin θ or y = cos θ . he period of y
= sin k θ or
y
= cos k θ is
3-# , k O #. he period of y k
= tan k θ is
1 . &e can !se k
1# k
, k O #.
he val!e of b determines the horiKontal translation, 'nown as the p#ase s#ift. he val!e of d determines the vertical translation. y = d is the e%!ation of the axis of t#e crve . e.g. e.g. y
= cos 2θ + 1 2
y
=
1 2
sin (θ + )*
)
1
g(x) = cos(2⋅x)+1
1.5 0.5
g(x) = 0.5⋅sin(x+45)
1
0.5 50
50
100
150
200
250
300
350
400
100
150
200
250
300
350
-0.5
-0.5
-1
f(x) = cos(x)
f(x) = sin(x) -1
400
