Sketching Curves

INTRODUCTION

• THIS CHAPTER FOCUSES ON SKETCHING GRAPHS

• WE WILL ALSO BE LOOKING AT USING THEM TO SOLVE EQUATIONS

• THERE WILL ALSO BE SOME WORK ON GRAPH TRANSFORMATIONS

SKETCHING CURVESSKETCHING CUBICS

YOU NEED TO BE ABLE TO SKETCH EQUATIONS OF THE FORM:

THIS INVOLVES FINDING THE PLACES WHERE THE GRAPH CROSSES THE AXES, IN THE SAME WAY YOU DO WHEN SKETCHING A QUADRATIC.

3 2y ax bx cx d

( )( )( )y x a x b x c

A cubic equation will take one of the following shapes

For any x3

For any -x3

3 2y ax bx cx d

( )( )( )y x a x b x c

ExampleSketch the graph of the function:

( 2)( 1)( 1)y x x x

If y = 00 ( 2)( 1)( 1)x x x

So x = 2, 1 or -1(-1,0) (1,0) and (2,0)

If x = 0(0 2)(0 1)(0 1)y

So y = 2(0,2)

3 2y ax bx cx d

( )( )( )y x a x b x c

( 2)( 1)( 1)y x x x

(-1,0) (1,0) (2,0) (0,2)

If we substitute in x = 3, we get a value of y = 8. The curve must be increasing after this

point…

3 2y ax bx cx d

( )( )( )y x a x b x c

( 2)(1 )(1 )y x x x

If y = 00 ( 2)(1 )(1 )x x x

So x = 2, 1 or -1(-1,0) (1,0) and (2,0)

If x = 0(0 2)(1 0)(1 0)y

So y = -2(0,-2)

3 2y ax bx cx d

( )( )( )y x a x b x c

( 2)(1 )(1 )y x x x

(-1,0) (1,0) (2,0) (0,-2)

If we substitute in x = 3, we get a value of y = -8. The curve must be decreasing after this

point…

3 2y ax bx cx d

( )( )( )y x a x b x c

2( 1) ( 1)y x x

If y = 020 ( 1) ( 1)x x

So x = 1 or -1(-1,0) and (1,0)

If x = 02(0 1) (0 1)y

So y = 1(0,1)

3 2y ax bx cx d

( )( )( )y x a x b x c

2( 1) ( 1)y x x

(-1,0) (1,0) (0,1)

point…

‘repeated root’

3 2y ax bx cx d

( )( )( )y x a x b x c

3 22 3y x x x

If y = 00 ( 3)( 1)x x x So x = 0, 3 or -1(0,0) (3,0) and (-1,0) If x = 0

0(0 3)(0 1)y So y = 0(0,0)

2( 2 3)y x x x

( 3)( 1)y x x x

FactoriseFactorise

3 2y ax bx cx d

( )( )( )y x a x b x c

3 22 3y x x x

(0,0) (3,0) (-1,0)

point…

SKETCHING CURVES

SKETCHING CUBICSYOU NEED TO BE ABLE TO SKETCH AND INTERPRET CUBICS THAT ARE VARIATIONS OF Y = X3

THIS WILL BE COVERED IN MORE DETAIL IN C2. YOU CAN STILL PLOT THE GRAPHS IN THE SAME WAY WE HAVE SEEN BEFORE. THIS TOPIC IS OFFERING A ‘SHORTCUT’ IF YOU CAN UNDERSTAND IT.

yy = x3

SKETCHING CURVES

yy = x3

y = -x3

A cubic with a negative ‘x3’ will be reflected in the x-axis

‘Whatever you get for x3, you now have the negative of

that..’

SKETCHING CURVES

3( 1)y x

yy = x3

When a value ‘a’ is added to a cubic, inside a bracket, it is a

horizontal shift of ‘-a’‘I will now get the same values for y, but with values of x that

are 1 less than before’

y = (x + 1)3

When x = 0:

3(0 1)y 1y

y-intercept

SKETCHING CURVES

3(3 )y x

yy = x3

y = (3 - x)3

When x = 0:

3(3 0)y 27y

y-intercept

3(3 )y x 3( 3)y x

Reflected in the x-

Horizontal shift, 3 to the right

SKETCHING CURVESTHE RECIPROCAL FUNCTION

YOU NEED TO BE ABLE TO SKETCH THE ‘RECIPROCAL’ FUNCTION. THIS TAKES THE FORM:

WHERE ‘K’ IS A CONSTANT.

ExampleSketch the graph of the function 1y

and its asymptotes.

124-4-2-1y

10.50.25-0.25-0.5-1x x

y = 1/x

You cannot divide by 0, so you get no value at this point

These are where the

graph ‘never reaches’, in this case the

axes…

and its asymptotes.

y = 1/x

y = 3/x

The curve will be the same, but further out…

and its asymptotes.

y = 1/x

y = -1/x

The curve will be the same, but reflected in

the x-axis

SKETCHING CURVESSOLVING EQUATIONS AND SKETCHING

YOU NEED TO BE ABLE TO SKETCH 2 EQUATIONS ON A SET OF AXES, AS WELL AS SOLVE EQUATIONS BASED ON GRAPHS.

ExampleOn the same diagram, sketch the

following curves:

( 3)y x x 2 (1 )y x x and

( 3)y x x Quadratic ‘U’ shapeCrosses through 0 and 3 0 3

( 3)y x x

2 (1 )y x x Cubic ‘negative’ shapeCrosses through 0

and 1. The ‘0’ is repeated so just

‘touched’

2 (1 )y x x

following curves:( 3)y x x 2 (1 )y x x an

dFind the co-ordinates of the points of intersection

These will be where the graphs are equal…

( 3)y x x

2 (1 )y x x

2( 3) (1 )x x x x 2 2 33x x x x 3 3 0x x 2( 3) 0x x

Expand bracketsGroup

togetherFactorise

0x 2 3 0x 2 3x

following curves:( 3)y x x 2 (1 )y x x an

dFind the co-ordinates of the points of intersection

These will be where the graphs are equal…2( 3) (1 )x x x x

2 2 33x x x x 3 3 0x x 2( 3) 0x x

Expand bracketsGroup

togetherFactorise

0x 2 3 0x 2 3x

( 3)y x x

x=-√3 x=0 x=√3( 3)y x x ( 3)y x x ( 3)y x x

3( 3 3)y 0(0 3)y

0y 3 3 3y

3( 3 3)y

3 3 3y

(0,0)(-√3 , 3+3√3) (√3 , 3-3√3)

following curves:2 ( 1)y x x

2 ( 1)y x x Cubic ‘positive’ shapeCrosses through 0

and 1. The ‘0’ is repeated. 0

Reciprocal ‘positive’ shape

Does not cross any axes

y = 2/x

YOU NEED TO BE ABLE TO SKETCH 2 EQUATIONS ON A SET OF AXES, AS WELL AS SOLVE EQUATIONS BASED ON GRAPHS.HOW DOES THE GRAPH SHOW THERE ARE 2 SOLUTIONS TO THE EQUATION..

following curves:2 ( 1)y x x

y = 2/x

2 2( 1) 0x xx

2 2( 1)x xx

2 2( 1) 0x xx

Set equations equal, and re-

arrange

And they cross in 2 places…

SKETCHING CURVESMORE TRANSFORMATIONS

YOU HAVE SEEN THAT A CURVE WITH THE FOLLOWING FUNCTION:

WILL BE TRANSFORMED HORIZONTALLY ‘-A’ UNITS.

A CURVE WITH THIS FUNCTION:

WILL BE TRANSFORMED VERTICALLY ‘A’ UNITS

( )f x a

f(x)f(x + 2)

f(x) + 2

2 units left

2 units up

f(x + 2) The x values reduce by 2 for the same y values

f(x) + 2 The y values from the original function increase by 2

SKETCH THE FOLLOWING FUNCTIONS:

F(X) = X2

STANDARD CURVE LABEL KNOWN POINTS

G(X) = (X + 3)2

MOVED 3 UNITS LEFT WORK OUT NEW ‘KEY POINTS’

H(X) = X2 + 3 MOVED 3 UNITS UP WORK OUT NEW ‘KEY POINTS’

y f(x)

y g(x)

y h(x)

GIVEN THAT:

I) F(X) = X3

SKETCH THE CURVE WHERE Y = F(X - 1). STATE ANY LOCATIONS WHERE THE GRAPHS CROSSES THE AXES.

F(X) = X3

F(X – 1) = (X – 1)3

SO FOR THIS CURVE, WHEN X = 0, Y = -1IT THEREFORE CROSSES AT Y = -1

f(x – 1)

GIVEN THAT:

I) G(X) = X(X – 2)

SKETCH THE CURVE WHERE Y = G(X + 1). STATE ANY LOCATIONS WHERE THE GRAPHS CROSSES THE AXES.

G(X) IS A POSITIVE QUADRATIC CROSSING AT 0 AND 2.

G(X) = X(X – 2)

G(X + 1) = (X + 1)(X + 1 – 2) G(X + 1) = (X + 1)(X – 1)

SO FOR THIS CURVE, WHEN X = 0, Y = -1IT THEREFORE CROSSES AT Y = -1

g(x + 1)

x’s replaced with ‘x + 1’

GIVEN THAT:I) H(X) = 1/X

SKETCH THE CURVE WHERE Y = H(X) + 1. STATE ANY LOCATIONS WHERE THE GRAPHS CROSSES THE AXES AND THE EQUATIONS OF ANY ASYMPTOTES.

H(X) IS A POSITIVE RECIPROCAL GRAPH

H(X) = 1/X

H(X) + 1 = 1/X + 1

THE ASYMPTOTES ARE: X = 0 (THE Y-AXIS)Y = 1

IT WILL CROSS THE X-AXIS AT -1 SINCE THIS VALUE WILL MAKE THE EQUATION = 0

h(x) + 1

SKETCHING CURVESEVEN MORE TRANSFORMATIONS

YOU ALSO NEED TO BE ABLE TO PERFORM TRANSFORMATIONS OF THE FORM:

THIS IS A HORIZONTAL STRETCH OF 1/A.

YOU ALSO NEED TO KNOW:

THIS IS A VERTICAL STRETCH BY FACTOR ‘A’

( )f ax

( )af x

(2 )y f x( )y f x‘We will get the same y values, using half the x

values’ This is because the x

values get multiplied by 2 before the y values are

worked out

2 ( )y f x( )y f x‘We will get y values twice as big, using the same x

values’ This is because when we

work out the y values, they are doubled after

GIVEN THAT F(X) = 9 – X2, SKETCH THE CURVE WITH EQUATION;A) Y = F(2X)

SKETCH THE ORIGINAL CURVE, WORKING OUT KEY POINTS.IF X = 0

IF Y = 0

y f(x)

(3 )(3 )y x x

0 (3 )(3 )x x

(3,0) (-3,0)

GIVEN THAT F(X) = 9 – X2, SKETCH THE CURVE WITH EQUATION;A) Y = F(2X)

SUBSTITUTE ‘2X’ IN PLACE OF ‘X’IF X = 0

IF Y = 0

y f(x)

y f(2x)

-1.5 1.5

929 (2 )y x

(3 2 )(3 2 )y x x

29 4y x

0 (3 2 )(3 2 )x x

(-1.5,0)

(1.5,0)

29 4y x

SKETCHING CURVESEVEN MORE TRANSFORMATIONSGIVEN THAT F(X) = 9 – X2, SKETCH THE CURVE WITH EQUATION;A) Y = 2F(X)

F(X), THE ORIGINAL EQUATION, IS DOUBLED..IF X = 0

IF Y = 0

y f(x)

y 2f(x)

1829y x

2(3 )(3 )y x x

22(9 )y x

0 2(3 )(3 )x x

(0,18)

(3,0) (-3,0)

22(9 )y x

SUMMARY

• WE HAVE LEARNT THE SHAPES OF SEVERAL DIFFERENT CURVES

• WE HAVE LEARNT HOW TO APPLY TRANSFORMATIONS TO THOSE CURVES

• WE HAVE ALSO LOOKED AT HOW TO WORK OUT THE ‘KEY POINTS’

Sketching Curves

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