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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
A cubic equation will take one of the following shapes
For any x3
For any -x3
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
( 2)( 1)( 1)y x x x
If y = 0
0 ( 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)
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
( 2)( 1)( 1)y x x x
(-1,0) (1,0) (2,0) (0,2)
x
y
2
2
-1
1
If we substitute in x = 3, we get a value of y = 8. The curve must be increasing after this
point…
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
( 2)(1 )(1 )y x x x
If y = 0
0 ( 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)
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
( 2)(1 )(1 )y x x x
(-1,0) (1,0) (2,0) (0,-2)
x
y
2
-2
-1
1
If we substitute in x = 3, we get a value of y = -8. The curve must be decreasing after this
point…
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
2( 1) ( 1)y x x
If y = 0
20 ( 1) ( 1)x x So x = 1 or -1
(-1,0) and (1,0)
If x = 0
2(0 1) (0 1)y So y = 1
(0,1)
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
2( 1) ( 1)y x x
(-1,0) (1,0) (0,1)
x
y
1
-1
1
If we substitute in x = 2, we get a value of y = 3. The curve must be increasing after this
point…
‘repeated root’
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
3 22 3y x x x
If y = 0
0 ( 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
Factorise
Factorise fully
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.
4A
3 2y ax bx cx d
( )( )( )y x a x b x c
or
ExampleSketch the graph of the function:
3 22 3y x x x
(0,0) (3,0) (-1,0)
x
y
0-1
3
If we substitute in x = 4, we get a value of y = 20. The curve must be increasing after this
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.
4B
ExampleSketch the graph of the function:
3y x
x
yy = x3
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.
4B
ExampleSketch the graph of the function:
3y x
x
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..’
5
-5
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.
4B
ExampleSketch the graph of the function:
3( 1)y x
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
1
When x = 0:
3(0 1)y 1y
y-intercept
-1
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.
4B
ExampleSketch the graph of the function:
3(3 )y x
x
yy = x3
y = (3 - x)3
27
When x = 0:
3(3 0)y 27y
y-intercept
3(3 )y x 3( 3)y x
Reflected in the x-
axis
Horizontal shift, 3 to the right
3
Sketching CurvesThe Reciprocal Function
You need to be able to sketch the ‘reciprocal’ function. This takes the form:
Where ‘k’ is a constant.
ky
x
ExampleSketch the graph of the function 1
yx
and its asymptotes.
124-4-2-1y
10.50.25-0.25-0.5-1x x
y
y = 1/x
4C
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…
Sketching CurvesThe Reciprocal Function
You need to be able to sketch the ‘reciprocal’ function. This takes the form:
Where ‘k’ is a constant.
ky
x
ExampleSketch the graph of the function 3
yx
and its asymptotes.
x
y
y = 1/x
4C
y = 3/x
The curve will be the same, but further out…
Sketching CurvesThe Reciprocal Function
You need to be able to sketch the ‘reciprocal’ function. This takes the form:
Where ‘k’ is a constant.
ky
x
ExampleSketch the graph of the function 1
yx
and its asymptotes.
x
y
y = 1/x
4C
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:
4D
( 3)y x x 2 (1 )y x x and
x
y
( 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’
1
2 (1 )y x x
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:
4D
( 3)y x x 2 (1 )y x x andFind the co-ordinates of the points of
intersection
These will be where the graphs are equal…
x
y
0 3
( 3)y x x
1
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
together
Factorise
0x 2 3 0x 2 3x
3x
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:
4D
( 3)y x x 2 (1 )y x x andFind 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
together
Factorise
0x 2 3 0x 2 3x
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)
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:
4D
2 ( 1)y x x 2
yx
and
x
y
2 ( 1)y x x Cubic ‘positive’ shapeCrosses through 0
and 1. The ‘0’ is repeated.
02
yx
Reciprocal ‘positive’ shape
Does not cross any axes
1
y = 2/x
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.
How does the graph show there are 2 solutions to the equation..
ExampleOn the same diagram, sketch the
following curves:
4D
2 ( 1)y x x 2
yx
and
x
y
0 1
y = 2/x2 2( 1) 0x x
x
2 2( 1)x x
x
2 2( 1) 0x x
x
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
4E
( )f x a
( )f x a
f(x)f(x + 2)
f(x) + 2
x
y
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
Sketching CurvesMore Transformations
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’
4E
x
y f(x)
0
x
y g(x)
-3x
y h(x)
3
9
Sketching CurvesMore Transformations
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
4E
f(x)
0x
y
f(x – 1)
1-1
Sketching CurvesMore Transformations
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
4E
g(x)
0x
y
1-1 2
g(x + 1)
x’s replaced with ‘x + 1’
-1
Sketching CurvesMore Transformations
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
4E
h(x)
x
y
1
-1
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’
4F
( )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
Sketching CurvesEven more Transformations
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
4F
x
y f(x)
-3 3
9
29y x
(3 )(3 )y x x
9y
29y x
0 (3 )(3 )x x
(0,9)
(3,0) (-3,0)
Sketching CurvesEven more Transformations
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
4F
x
y f(x)
-3 3
9
x
y f(2x)
-1.5 1.5
929 (2 )y x
(3 2 )(3 2 )y x x
9y
29 4y x
0 (3 2 )(3 2 )x x
(0,9)
(-1.5,0)
(1.5,0)
29 4y x
Sketching CurvesEven more Transformations
Given 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
4F
x
y f(x)
-3 3
9
x
y 2f(x)
-3 3
18
29y x
2(3 )(3 )y x x
18y
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’
Top Related