Math - Refresher Course (graphing)

Refresher Topic : GRAPHING TECHNIQUES CURRICULUM OBJECTIVES ¾

Use of a graphic calculator to graph a given function

¾

Relating the following equations with their graphs

¾ ¾

Characteristics of graphs such as symmetry, intersections with the axes, turning points and asymptotes Determining the equations of asymptotes, axes of symmetry, and restrictions on the possible values of x and/or y

¾

¾ ¾

ax + b ax 2 + bx + c x2 y 2 y = y ± = 1 , = , dx + e cx + d a 2 b2

y = af ( x) , y = f ( x) + a , y = f ( x + a) , y = f (ax) , and combinations of these transformations eg . y = f (ax + b) 1 and y 2 = f ( x) and y = f ′( x) to the graph of y = f ( x) Relating graphs of y = f ( x ) , y = f ( x ) , y = f ( x)

Effect of transformations on the graph of

y = f ( x)

as represented by

Simple parametric equations and their graphs

GRAPHS OF PARABOLA Parabolas given by the equation

y 2 = kx (k > 0) are symmetrical about the x-axis and those given by x2 = ky

(k > 0) are symmetrical about the y-axis.

y 2 = kx

y

x 2 = ky

x

GRAPHS OF ELLIPSES 2 2 Graphs of the form ( x − h ) + ( y − k ) = 1 are ellipses with coordinates of centre (h, k). a2 b2 y These ellipses are symmetrical about the lines x = h and y = k . b

a

y b

(h, k)

a

(h, k) x

a>b

Special case: a = b

x

a
Circles are special cases of ellipses, just as squares are special cases of rectangles. When a = b = r, equation of the ellipse becomes ( x − h )2 + ( y − k )2 = r 2 which is the equation of a circle centered at (h, k) with radius r. Alternatively, it can be given in the form x 2 + y 2 + 2 cx + 2 dy + e = 0 . GRAPHS OF HYPERBOLAS [refer eg 1] 2 2 2 2 Graphs of the form x − y = 1 or y − x = 1 are hyperbolas. These hyperbolas are symmetrical about both c2 d 2 d 2 c2 axes and the lines y = ± d x are the asymptotes of the graphs. y 2 x2 c x2 y 2 − =1 d − =1 y = x d 2 c2 c2 d 2 c d

y=

−c

d

c d y=− x c

Axis/Point of

Graphically

Symmetry

y-axis x-axis origin

−d

y=−

c

x

d x c

Algebraically

If (x, y) is a point on the graph of

Equation of the curve remains unchanged when x is

y = f(x) then (−x, y) is also a point on the same graph.

replaced by –x. eg.

If (x, y) is a point on the graph of

Equation of the curve remains unchanged when y is

y = f(x) then (x, −y) is also a point on the same graph.

replaced by −y.

If (x, y) is a point on the graph of y = f(x) then

Equation of the curve remains unchanged when x is

(−x, −y) is also a point on the same graph.

replaced by –x and y by –y.

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eg.

y = x4 y2 = x eg. y = x 3 , y =

1 x

Transformation

Description of transformation

y = f ( x ) → y = −f ( x )

The graph of

Geometrically

( x , y) → ( x , − y)

y = −f ( x) is the reflection of the graph of

y = f ( x) in the x-axis. y = f ( x) → y = f (− x)

( x , y) → (−x , y)

The graph of y = f ( − x ) is the reflection of the graph of

y = f ( x) in the y-axis. y = f ( x) → y = f ( x) + a

( x , y) → ( x , y + a)

The graph of y = f ( x) + a is the translation of the graph of

y = f ( x) by a units in the positive y-direction.

y = f ( x) → y = f ( x) − a

( x , y) → ( x , y − a)

The graph of y = f ( x) − a is the translation of the graph of

y = f ( x) by a units in the negative y-direction. y = f ( x) → y = f ( x + a ) y = f ( x) → y = f ( x − a ) y = f ( x) → y = af ( x) y = f ( x) → y = f (ax)

The graph of

( x , y) → ( x − a , y)

y = f ( x + a) is the translation of the graph of

y = f ( x) by a units in the negative x-direction. The graph of y = f ( x − a ) is the translation of the graph of y = f ( x) by a units in the positive x-direction. The graph of y = af ( x) is the scaling of the graph of y = f ( x) by factor a parallel to the y-axis.

( x , y) → ( x + a , y) ( x , y ) → ( x , ay )

The graph of y = f (ax ) is the scaling of the graph of y = f ( x) by factor 1 parallel to the x-axis. a

Modulus:

⎧ f ( x) , f ( x) ≥ 0 | f ( x ) |= ⎨ ⎩ −f ( x ) , f ( x ) < 0

( x , y ) → ⎛⎜

x , ⎝a

⎧ f ( x) , x ≥ 0 f (| x |) = ⎨ ⎩f ( − x ) , x < 0

Composition of Two transformations: [refer to eg 6] y = f (kx + a) is a composition of y = f ( x + a) followed by y = f (kx) ; • or

•

a composition of y = f (kx) followed by y = f ( x + ak ) .

y = mf ( x) + b is a composition of y = mf ( x) followed by y = f ( x ) + b .

SYSTEMATIC CURVE SKETCHING (1) Intercepts on Coordinate Axes (2) Linear Asymptotes : Vertical / Horizontal & Oblique Asymptotes (3) Stationary Points •

First Derivative Test Given dy = 0 at point where x = a. We test the value of dy at points around x = a. dx dx

x

a−

a

a+

x

a−

a

a+

dy dx

+

0

−

dy dx

−

0

+

Shape

Shape

( a, f ( a ) ) is a maximum point. x

a−

a

a+

x

a−

a

a+

dy dx

+

0

+

dy dx

−

0

−

Shape

Note:

( a, f ( a ) ) is a minimum point.

Shape

( a, f ( a ) ) is a stationary point of inflexion.

If dy > 0, then f is strictly increasing, i.e., dx

y

If dy < 0, then f is strictly decreasing, i.e., dx http://education.helixated.com An Open Source Education Project

increases as

y

x

decreases as

increases.

x

increases.

⎞ y⎟ ⎠

•

Second Derivative Test

dy =0 dx

If

at x = a and Or Or

d2y > 0, then ( a, f ( a ) ) is a minimum point. dx 2 d2y < 0, then ( a, f ( a ) ) is a maximum point. dx 2 d2y = 0, no conclusion can be drawn, so use 1st derivative test dx 2

GRAPHS OF RATIONAL FUNCTIONS y = Given a rational function f ( x ) = P ( x ) , we express Q (x)

ax + b ax 2 + bx + c and y = , a≠0 dx + e cx + d

f ( x)

where degree of R(x) < degree of Q(x). Then, the horizontal/oblique asymptote is values of x where Q(x) = 0 and

y = f ( x)

y = ± f ( x) [refer Eg 6 for y 2 = f ( x) and y 2 = −f ( x) ]

can be obtained from

y = f ( x) , by considering the following features:

The graph is defined only for values of x on the graph of

Graph of

y = f ( x)

where

f ( x) ≥ 0 . Graph of y =

y = f ( x)

f ( x)

increases.

f ( x)

increases.

f ( x)

decreases.

f ( x)

decreases.

let

let

b > 0:

f ( x)

( a, b ) is a maximum point on

y = f ( x) .

( a, b ) is a maximum point on y =

f ( x) .

( a, b ) is a minimum point on

y = f ( x) .

( a , b ) is a minimum point on y =

f ( x) .

b > 0 : y = f ( x)

has a horizontal asymptote at

y = f ( x) Note:

y = g ( x) , and the vertical asymptotes are

R( x) ≠ 0 .

GRAPH OF y 2 = f ( x) ⇒ Graph

R (x) , Q (x)

f (x) = g(x) +

as a proper fraction, i.e.,

has a vertical asymptote at

y = b.

x = a.

y = f ( x) has a horizontal asymptote at y = b . y = f ( x)

y 2 = f ( x) ⇒ y = ± f ( x) . Hence the curve y 2 = f ( x)

has a vertical asymptote at

x = a.

is symmetrical about the x-axis.

GRAPH OF y = 1

f(x)

Graph of

Graph of y = 1 f(x)

y = f ( x)

f ( x)

increases.

1 decreases. f(x)

f ( x)

decreases.

1 increases. f(x)

( a, b ) is a maximum point. ( a, b ) is a minimum point.

let b ≠ 0:

y = f ( x)

has an x-intercept at

y = f ( x)

has a vertical asymptote at

x = a. x = a.

( a, b1 ) is a minimum point. ( a, b1 ) is a maximum point. y= 1

has a vertical asymptote at

y= 1

has an x-intercept at

f(x) f(x)

x = a.

x = a.

CURVES DEFINED BY PARAMETRIC EQUATIONS At times it is more convenient to describe a curve using a parameter. The pair of equations for parameter

t

are called parametric equations of the curve, i.e., we may write

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x = g(t )

and

x and y given in terms y = h(t ) instead of y = f ( x) .

of a

Eg 1. Sketch ( x − 1) 2 22

−

( y + 1) 2 2

=1

2

and find the equations of the asymptotes.

Eg 2. 2008 J1 Term Exam Q10 A curve has parametric equations x

= cos3 2t , y = sin 3 2t , for 0 ≤ t ≤

π 4

. (i)

(ii) Find the equations of the tangent and the normal to the curve at the point P

Sketch the curve.

( cos

3

[1]

2θ ,sin 3 2θ ) , where 0 ≤ θ ≤

π 4

(iii) The tangent at the point P meets the x- and y-axes at S and T respectively. Show that distance ST is a constant.

.[4] [3]

Eg 3. SAJC07/I/7 The curve C has equation y =

a x 2 + 2 where x ≠ 1 and a is a non-zero constant . x − 1

Show that if C has no stationary points, then −2 < a < 0 .

* (i) (ii)

[3]

It is given that the line y = – x – 1 is an asymptote of C. Find the value of a.

[2]

(iii) Sketch C, showing clearly asymptotes & coordinates of any intersections with the axes.

[3]

Eg 4 SRJC2007/I/10 modified The curve C has equation y =

y-axis at –2.

ax 2 + bx + c . It is known that C, that has asymptotes x = 3 and y = x − 1 , cuts the x+d

(i) (ii)

Find the values of a, b, c and d. [a = 1 , b = - 4 , c = 6 , d = - 3] [4] * Sketch the curve C, stating clearly the turning points, asymptotes and any intercepts with the axes. Hence deduce the 2 range of values of k such that the equation ax + (b − k ) x + c − kd = 0 has two distinct real roots. [5]

(iii)

Sketch graph of y =

ax 2 + b x + c stating clearly turning points, asymptotes and any intercepts with axes [3] x +d

2 Eg 5. The curve has equation f ( x ) = ( x − 9 ) . Sketch the curve if (i) C = 0 (ii) C < 0 (iii) C > 0 and state the equations of all the 2

x +c

asymptotes of C and the coordinates of where C cuts the axes. Eg 6 YJC 2007/I/9 (modified) The curve shown in the diagram has equation and has a maximum point at

1⎞ ⎛ ⎜ 3.5, ⎟ . 9⎠ ⎝

y = f (x ) . It intersects the axes at x = 3, y

x = 4, y = 3

3 1 ⎛ ⎜ 3 .5 , 9 ⎝

3

0

On separate diagrams, sketch the graphs of: (i)

y = f ( 2 x − 1) ,

(ii)

y = f ( x) 2

What should you do if you need to sketch http://education.helixated.com An Open Source Education Project

x=2 (iii)

y=

y = − f (x ) , 2

1 f ( x + 1)

?

⎞ ⎟ ⎠

y=2

4

x

x=5 (iv)

y=

1 . f ( x)

y

Eg 7 NJC2007 /I/6

y = f ( x)

−3

−1

x

−2 −1

A (2, −2)

Sketch, on separate diagrams, the following graphs, indicating clearly any asymptotes, axial intercepts and turning

y = −f ( x ) − 1 ;

points, where possible. (i)

*(ii)

y = f (− x ) ;

y = f '( x) .

*(iii)

[6]

Self Practice TPJC 2007/I/8 (modified) On separate diagrams sketch the following graphs showing the axial intercepts, stationary point and equations of asymptotes. a)

y=

x (3 x − 10) ( x − 3)

b)

2

x (3 x − 10)

y=

( x − 3)

c)

2

y2 =

x (3x − 10) ( x − 3) 2

[6]

CJC 2007/I/12 (modified) The curve C has equation

f ( x) =

Q R ( x − 9) 2 . Express f(x) in the form P + , and state the + ( x − 3)( x + 3) x −3 x +3

equations of all the asymptotes of C and the coordinates of where C cuts the axes. Sketch on separate diagrams, (a)

y = f ( x) ,

(b)

1 y= , f ( x)

[6]

y = − f (2 x − 1)

(c)

[9]

making clear the main relevant features of each curve. HCI 07/I/13

The curve

It is given that

C

C

has equation

x 2 + ax + 4 . x+b

has a vertical asymptote

x = −1

and a stationary point at

x = 2.

(i)

Determine the values of

(ii)

Find the equation of the other asymptote of

(iii) (iv) (v)

Prove, using an algebraic method, that C cannot lie between two values (to be determined). [4] Draw a sketch of C, showing clearly any axial intercepts, asymptotes and stationary points. [3] Deduce the number of real roots of the equation

a

and

b.

( 4 − x ) ( x + 1) = ( x 2

2

ans:

y=

1) y + 1 = ±0.707( x − 1)

(

2

[3]

− 4x + 4

3) a = – 1

)

C.

)

2

[1]

.

[2]

4) k > 5.46 or k < −1.46

(

)

2) y − sin 3 2θ = − tan 2θ x − cos3 2θ ; y − sin 3 2θ = cot 2θ x − cos3 2θ ; ST = 1 5) (ii) asym : x = ± c , y= 1 (iii) asym y= 1 6) (i) (0.5,3) (2.25, 0.111) (2,0) (2.5,0) y = 2, x = 1.5 , x = 3 (ii) (3.5,0.333) (0, 0.111) (0,1.73) (0, -1.73) y = 1.41, y = -1.41, x = 2 , x = 5 (iii) (3,0) (4,0) x = 2 , x = 5 (iv) (0,0.333) (2,0) (5,0) (3.5,9) x = 3, x =4 , y = 0.5 7) (i) x = -2 , y = 0 , A(2,1) (ii) (-3,0) (-1,0)(1,0)(3,0) (0,-1) x = -2, x= 2, y = -1 (iii) x = -2, y = 0, (2,0) HCI : a = -4 , y = x -5, C cannot lie between -12 and 0; (0,4) (2,0) (-4,-12) y=x-5, x =-1; 2 roots TP: (a) y=3, x=3, (0,10/3),(0,0) (b) y=3, x=3, (0,10/3),(0,0) (c) y = ± 3 , x=3, (3.75,±2.89)(0,0)

6 24 , x = 3 , x = -3, y = 1 min(9,0) max (1,-8) ; (-3,0) (3,0) x =9,y=1, min(1, -1/8) − ( x − 3) ( x + 3) min(1.45, 8.065), max(5,0), y = -1, x = -1, x=2

CJ: = 1 +

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Refresher Topic:

FUNCTIONS

CURRICULUM OBJECTIVES

¾ Concept of function, domain and range ¾ Use of notations such as f ( x) = x 2 + 5, f : x 6 x 2 + 5, f −1 ( x), fg( x) and f 2 ( x) ¾ ¾ ¾ ¾

Finding inverse functions and composite functions Conditions for the existence of inverse functions and composite functions Domain restriction to obtain an inverse function Relationship between a function and its inverse as reflection in the line y = x

1. ONE-ONE FUNCTIONS

A function f is one to one if no two elements in its domain have the same image. We can prove that a function is one-one analytically or graphically. Analytical Method: If we can prove that for any x1 , x2 ∈ Df : x1 ≠ x2 ⇒ f ( x1 ) ≠ f ( x2 ) f ( x1 ) = f ( x2 ) ⇒ x1 = x2

or equivalently,

Graphical Method (Horizontal line test): The horizontal line y = k , for all k ∈ Rf , cuts the graph of y = f ( x ) once and only once, then f is a one-one function.[Note: The graph of y = f ( x ) must be sketched when using horizontal line test.] 2. INVERSE FUNCTIONS

The inverse of function f is such that if y = f ( x ) , then x = f −1 ( y ) . The inverse function of f (denoted by f −1 ) exists if and only if f is one-one. Notes: • Domain of inverse function f −1 , Df -1 = Rf , Range of function f.

•

Range of inverse function f −1 , Rf -1 = Df , Domain of function f.

•

Graphically, the graphs of

y = f ( x)

and

y = f −1 ( x )

are reflections of each other

in the line y = x. 3. COMPOSITE FUNCTIONS

Let f and g be functions. gf is defined as gf ( x ) = g ⎡⎣ f ( x ) ⎤⎦ .

f

Important Notes:

1. Dgf = Df .

Dg

Df

Similarly, Dfg = Dg .

Rg g

Rf

Rg f

2. gf may or may not be a function. gf is a composite function if and only if Rf ⊆ Dg. Similarly, fg is a composite function if and only if Rg ⊆ Df.

gf

3. To find the range of the function gf we can either •

sketch the graph of gf and read off its range or

•

if the graphs of g and f are available (and especially if the graph of gf is very difficult to sketch), then we can use the range of f as the restricted domain of g and read off the corresponding range to find Rgf . i.e. Rgf = Set of images of Rf under g .

4. gf ≠ fg in general, i.e. composition of functions is not commutative. http://education.helixated.com An Open Source Education Project

HCI 07/I/12

f :x→

The functions f and g are defined by

( x + 3)( x − 1)

g:x →e

(i)

4

− 3− x

x ∈ \, x ≠ −3, x ≠ 1

,

x∈\

,

Sketch, on separate clearly labelled diagrams, the graphs of y = f ( x ) and y = g ( x ) . Hence, or otherwise, show that the inverse function of g does not exist.

(ii)

The function h is defined by

[3]

h : x → g ( x ) , x ∈ A , where A is the maximal subset of the

interval ( 0, ∞ ) such that the inverse function h −1 exists. State the set A and give the rule and domain of h −1 .

[4]

(iii) Determine which of the following is a function: a) f g b) f h -1 In each case, if the function exists, give its domain and range.

NJC 07/I/7 The functions f and g are given by

[3]

f : x 6 ( x − 2) 2 − 3, x ∈ R, x ≤ 2 g : x 6 a − e− x , x ∈ R

(i)

Show that f −1 exists and express f −1 in similar form, stating the domain clearly. [3]

(ii)

Determine the largest integer value a such that fg exists.

(iii)

For the largest value of a obtained in part (ii), determine the domain and range of fg. [2]

[2]

Self Practice Term Exam J1 2008 Q6

The functions f and g are defined by

f : x → ln( x + 1) g : x → ex

2

,

x > −1

,

x ∈ \−

The function gf is defined if the domain of f is restricted. State the largest possible set of [2] values of x for which gf is defined. (ii) Using the restricted domain of f, define gf and state its range. [3] (i)

PJC 07/I/10 The functions f and g are defined by f : x → x2 + 6x − 9 , x ∈ \− , 1 g:x → 2 , a > 0, x ≠ a, x ∈ \ , a − x2 h : x → 2 − 3x − 1, x > k . (i) Sketch, on separate diagrams, the graphs of y = f ( x ) and y = g ( x ) . (ii) (iii) (iv)

[2]

State the maximal domain of g for which its inverse function exists. [1] Find the range of values of a for which the equation h(x) – g(x) = 0 has two real roots, giving your answer in exact form. [3] Find the least value of k for which the function fh exists. [3]

HCI: [3, ∞) , h −1 ( x) = 3 − ln x, x ∈ (0,1] , f h -1 ( x) =

4 , x ∈ (0,1] , range 1/3 (6 − ln x)(2 − ln x)

NJ:

f −1 : x 6 2 − x + 3, x ∈ R, x ≥ −3 , a = 2, Dfg = Dg = \ , Rfg = ( −3, ∞ )

PJ:

(−∞, 0] or [0, ∞) , a >

11 5 , k= 18 3

Term Exam: (-1,0) gf ( x) = e[ln( x +1)] , x ∈ (−1, 0) , Rg f = (1, ∞ ) 2

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