The graph of the function y = x3 is shown at right: Both the domain and range of the function are R. The function is constantly increasing and has a stationary point of infl ection (where the gradient is 0) at the origin (0, 0).
Throughout this section we shall refer to the shape of the graph of y = x3 as a positive cubic, or a basic cubic curve.
Cubic functions are also power functions. Power functions are functions of the form f (ff x) = xn, n∈ R. The value of the power, n, determines the type of function. When n= 1, f (ff x) = x and the function xx
is linear. When n= 2, f (x) = x2 and the function is quadratic. When n= 3, f (x) = x3 and the function is cubic. When n= 4, f (x) = x4 and the function is quartic. Other power functions will be discussed later.
Under a sequence of transformations of f (x) = xn, n∈ R, the general form of a power function is f (ff x) = a(x − b)n+ c (where a, b, c and n ∈ R). All linear and quadratic polynomials are also linear and quadratic power functions, but this is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form f (ff x) = a(x − b)3+ c has exactly one x-intercept and one stationary point of infl ection. A cubic polynomial in the form f (
ff x) = ax3+ bx2+ cx + d can have one, two or three dd x-intercepts and is therefore not a power function.
All cubic power functions are also cubic polynomials, but not all cubic polynomials are cubic power functions. For example, the cubic function y= 2(x − 3)3 + 1 is a polynomial and a power function. It is the graph of y= x3 under a sequence of transformations.
Dilation
The value a is the dilation factor; it dilates the graph from the x-axis.
The larger a is, the thinner the graph.
Refl ection
If a is negative, the graph of the basic cubic is refl ected in the x-axis, that is, the graph is ‘fl ipped’ upside down.
If x is replaced with xx −x, the graph of the basic cubic is refl ected in the y-axis, that is, the graph is ‘fl ipped’ sideways.
For example, the graphs y= (x − 1)3 and y= (−x− 1)3 are refl ections of each other across the y-axis.
y eBookplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplus eBoo eBookkkplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplus
Digital doc Spreadsheet 015 Cubic function —
y a c
Translation
Horizontal translation
If b> 0, the graph of the basic cubic is translated horizontally to the right, and if b< 0, the graph of the basic cubic is translated horizontally to the left. For example, the graph with equation y= (x − 2)3 is a basic cubic translated 2 units to the right, and the graph of y= (x + 3)3 is a basic cubic, translated 3 units to the left, that is, parallel to the x-axis in the negative direction.
If the coeffi cient of x is not 1, the equation must be rewritten in xx the form y= a(x − b)3+ c in order to be able to work out the value of b. For example, the graph of y= (2x (2 (2 − 5)3 is translated 5
2 units to the right, since y = (2x (2(2 − 5)3
= −
= −
=[ ( −
= − )]
( )
( )
= −
= −
[ (
=[ ( −
= −
= −
= −
=8 −
= −
5 2 3
( 5) ( 2)
( )
( )3
= x−
= −
( )
( )
= −
= −
= −
= −
Vertical translation
The value of c translates the graph vertically or along the y-axis. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down.
The coordinates of the stationary point of infl ection are (b, c).
For example, if y= x3 is translated 1 unit up, the equation of the resulting graph is y= x3+ 1 and the point of infl ection is (0, 1); if it is translated 2 units down, the equation of the resulting graph is y= x3− 2 and the point of infl ection is (0, −2).
Combination of transformations
The graph of y= a(x − b)3+ c shows the combination of the transformations described above.
Finally, the domain and range of y = a(x − b)3+ c are R (all real numbers).
WORKED EXAMPLE 4
State the changes necessary to transform the graph of y= x3 into the graph of y= 2(x 2( 2( + 1)3− 4.
THINK WRITE
1 Write the general equation of the cubic function. y= a(x − b)3+ c
2 Identify the value of a. a= 2
3 State the effect of a on the graph. The graph is dilated by the factor of 2 in the y direction.
4 Identify the value of b. b=−1
5 State the effect of b on the graph. The graph is translated 1 unit to the left.
6 Identify the value of c. c =−4
7 State the effect of c on the graph. The graph is translated 4 units down.
y
0 x
−3 2
b=−3 b= 2
y= (x − b)3
y
0 x
−22 1
c= 1 c=−2
y= x3+ c
y
0 x
y= a(x − b)3+ c
(b, c)
WORKED EXAMPLE 5
For each of the following graphs:
i state the coordinates of the stationary point of infl ection ii fi nd the x- and y-intercepts iii sketch the graph
iv state the transformations that the graph of iv
iv y= x3 has undergone to form each new equation.
a y=−(x(( + 3)3− 1 b y= (4 − x)3+ 6
THINK WRITE/DRAW
a Write the equation. a y=−(x+ 3)3− 1
i Since the rule is of the form y= a(x − b)3+ c, identify the values of b and c and hence write the coordinates of the stationary point of infl ection (b, c).
i b=−3, c=−1
Stationary point of infl ection: (−3, −1)
ii Find the y-intercept by letting x= 0.
Find the x-intercept by letting y= 0. ii y-intercept: x= 0, y=−(0 + 3)3− 1 =−27 − 1 =−28
x-intercept: y= 0 (x+ 3)3− 1 = 0 (x+ 3)3=−1
x+ 3 =−1 x=−4 iii To sketch the graph on a set of labelled axes,
mark the stationary point of infl ection and the x- and y-intercepts, then sketch the positive cubic passing through the points marked.
iii
0 x
y
(−3, −1) (−4, 0)
iv State the kind of refl ection and the vertical and horizontal translations.
iv The graph is refl ected in the x-axis.
There is a horizontal translation of 3 units to the left and a vertical translation of 1 unit down.
b Write the equation. b y= (4 − x)3+ 6
i Since the rule is of the form y= a(x − b)3+ c, identify the values of b and c and hence write the coordinates of the stationary point of infl ection (b, c).
i b= 4, c = 6
Stationary point of infl ection: (4, 6)
ii Find the y-intercept by letting x= 0.
Find the x-intercept by letting y= 0.
Note: Do not round off until the very last step;
for graphing purposes, round off your fi nal answer to 1 decimal place.
ii y-intercept: x= 0, y= (4 − 0)3+ 6 = 64 + 6 = 70 x-intercept: y= 0 (4 − x)3+ 6 = 0 (4 − x)3=−6
44 66
4 6
4 6
4 36
4 6
4 36
4 6
4 6
4 6
= +44 66
4 − 6
4 6
4 6
4 6
4 6
4 6
x ≈ 5.8
iii To sketch the graph on a set of labelled
axes, mark the stationary point of infl ection and the x- and y-intercepts, then sketch the positive cubic passing through the points marked.
iii
0 x
y
(5.8, 0) (4, 6)
iv State the kind of refl ection and the vertical and horizontal translations.
iv The graph is refl ected in the y-axis.
There is a horizontal translation of 4 units to the right and a vertical translation of 6 units up.
To fi nd the equation of the curve from a given graph, we need to establish exactly what transformations were applied to the basic cubic curve. This is best done by observing the shape of the graph and the position of the stationary point of infl ection.
WORKED EXAMPLE 6
Find the equation of the curve, if it is of the form y= a(x(( − b)3+ c.
THINK WRITE/DISPLAY
1 Write the general equation of the cubic function. y= a(x − b)3+ c
2 Write the coordinates of the stationary point of infl ection (b, c) and hence state the values of b and c.
The stationary point of infl ection is (1, 3).
So b = 1, c = 3.
3 Substitute the values of b and c into the general formula.
y= a(x − 1)3+ 3
4 The graph passes through the point (0, 5) (y-intercept). Substitute the coordinates of this point into the equation.
Using (0, 5):
5 = a(0 – 1)3+ 3
5
6 Write the solution for the equation. a=−2
7 Substitute the value of a into y = a(x − 1)3+ 3. y=−2(x− 1)3+ 3
y
0 1 x 3 5
On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry line as:
solve(5 = a × (0 – 1)3 + 3,a) Then press ENTER · .
The graph of
1. y= x3 is a basic cubic curve.
The graph of
2. y = a(x − b)3+ c where a > 0 is the basic cubic, dilated by the factor of a from the x-axis, translated b units along the x-axis (to the right if b> 0, or to the left if b< 0) and c units along the y-axis (up if c > 0, or down if c < 0).
y
0 x
y= a(x − b)3+ c
(b, c)
If
3. a< 0, the graph is refl ected in the x-axis.
The stationary point of infl ection is at (
4. b, c).
Both the domain and range are
5. R.
REMEMBER
The cubic function in power form
1 WE4 State the changes necessary to transform the graph of State the changes necessary to transform the graph of y = x3 into the graph of each of the following.
a y= 7x3 b yyyyyy − 23xxxxxx3
c y= x3+ 4 d y= 6 − x3
e y= (x − 1)3 f y=−(x+ 3)3 g y= 4(2 − x)3 h y=−6(7 − x)3 i y= 3(x + 3)3− 2 j yyyyyyyyyyyyyyyyyyyyyyy 66666 12(xxxxxxxxxxxxxxxxxxxxxxx−−−−11111)3 k yyyyyyyyyyyyyyyyyyyyyyyyyyyy=========14(222xxxxxxxxxxxxxxxxxxxxxxxxxxxx+++++++++555)3 l yyyyyyyyyyyyyyyyyyyyyyyyyyy 3 2(44++1211xxxxxxxxxxxxxxxxxxxxxxxxxxx)3
2 Which of these transformations were applied to the graph of y = x3 to obtain each of the graphs below?
i refl ection in the x-axis ii translation to the left iii translation to the right
iv translation up v translation down
a y
0 x
b y
0 x
c y
0 x
d y
0 x
e y
0 x
f y
0 x EXERCISE
2B
3 WE5 For each of the following graphs:
i fi nd the stationary point of infl ection ii fi nd the x- and y-intercepts
iii state the transformation(s) that the graph y= x3 has undergone to produce the given graph
iv sketch the graph.
a yyyyyy 34xxxxxx3 b y= 1 − 2x 2 2 3
4 MC The coordinates of the stationary point of inflection are:
A ((−4,, )
B thinner and translated not as far to the right C shifted further to the left
D shifted further to the right E shifted further down
7 Find the equation of the graph resulting from each of the following transformations of the graph of y= x3:
a a dilation by the factor of 1
2 from the x-axis
b a refl ection in the x-axis and a translation by 5 units to the left c a translation by 3 units to the right and 1 unit down
d a dilation in the y direction by the factor of 2, followed by the vertical translation of 3 units
e a refl ection in the x-axis, then a translation of 1 unit to the left and 1 unit down.
8 Find the equation of the graph resulting from the following sequential transformations of the graph of y= x3:
a dilation by a factor of 2 from the x-axis b refl ection in the y-axis
c translation of 2 in the positive direction parallel to the x-axis d translation of 1 in the negative direction parallel to the y-axis.
9 WE6 Find the equations of these curves, if they are of the form y = a(x (( − b)3+ c. eBookplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplus eBoo eBookkkplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplusplus Spreadsheet 236 Spreadsheet 236 Function grapher EXAM TIP Be careful when sketching Be careful when sketching
graphs — use appropriate scales on the axes, clearly draw the graph and use the correct domain. Show at least one scale point on each axis and two coordinate points on each curve (including intercepts).
[Authors’ advice]
10 MC The graph of y = 2(x 2(2( + 3)3+ 1 has been reflected in the x-axis, shifted 3 units to the right and 1 unit up. The equation of the resulting graph is:
A y= 2(3 − x)3+ 2 B y=−2(x+ 2)3+ 2 C y= 2(2 − x)3 D y=−2(x− 3)3+ 1 E y=−2x22 3
11 The graph of a cubic function of the form y = a(x (( − b)3+ c has a stationary point of inflection at (−1, −4) and cuts the y-axis at y =−2. Find the equation of the function.
12 The graph of y = a(b − x)3+ c has a stationary point of inflection at (2, 1) and passes through the point ( , )( ,( ,12 .
a Find the equation of the curve.
b State the shape of the curve (that is, whether it is positive or negative cubic).
13 The graph of y = a(x (( − b)3+ c cuts the x-axis at x =−4 and the y-axis at y= 28. If it is known that the dilation factor is equal to 1:
a fi nd the position of the stationary point of infl ection b sketch the graph.