If a < 0, that is, each of the above rules is multiplied by −1, then the graphs are refl ected through the x-axis.
For example, the graph of y = −x4 (at right) is a refl ection, through the x-axis, of the graph of y = x4.
Similarly, the graph of y = −x4 + x2 = −(x4 − x2) is a refl ection through the x-axis of the graph of y = x4 − x2.
Note: The above graphs can be translated horizontally or vertically but this is considered in chapter 2.
To fi nd the x-intercepts of a quartic function, let y = 0 and solve the equation for x.
Repeated factors touch the x-axis as they do for cubic and quadratic functions.
Worked exAMple 27
Sketch the graph of y x== 4−−x3−−7x2++5x++10, showing all intercepts.
ThINk WrITe/drAW
Method 1: Technology-free
1 Find the y-intercept. When x = 0, y = 10
The y-intercept is 10.
2 Let y = P(x). Let P(x) = x4 − x3 − 7x2+ 5x + 10
3 Find two linear factors of the quartic expression, if
possible, using the factor theorem. P(1) = (1)4 − (1)3 − 7(1)2 + 5(1) + 10
= 8
≠ 0
P(−1) = (−1)4 − (−1)3 − 7(−1)2 + 5(−1) + 10 (x + 1) is a factor.= 0
P(2) = (2)4 − (2)3 − 7(2)2 + 5(2) + 10 (x − 2) is a factor.= 0
4 Find the product of the two linear factors. (x + 1)(x − 2) = x2 − x − 2
5 use long division to divide the quartic by the quadratic factor x2 − x − 2 (or use another method).
x2
)
−−− − − − + +
−( −
5
2 7 5 10
2 4 3 2
4
x x x x x x
x x33 2
2
2 0 5
−
− +
x x )
55 10 5 2 5 10
x
x x
+
−(− + + )
0
6 Express the quartic in factorised form. y = (x + 1)(x − 2)(x2 − 5)
7 Factorise the quadratic factor, x2 − 5, using difference of perfect squares.
y = (x + 1)(x − 2)(x + 5)(x − 5)
8 To fi nd the x-intercepts, set y equal to zero. Let y = (x + 1)(x − 2)(x + 5)(x − 5) = 0
9 Solve for x using the Null Factor Law. x = −1, 2, ± 5
10 State the x-intercepts. The x-intercepts are −1, 2, − 5 and 5.
y 0 x
y = −x4 y
0 x
y = −x4+ x2
eBookplus eBookplus
Tutorial int-0520 Worked example 27
11 Sketch the graph of the quartic. y
0 x 5, 0)
(− ( 5, 0)
1 2 3
(−1, 0)
(2, 0) (0, 10)
−3 −2 −1
Method 2: Technology-enabled
1
2
Worked exAMple 28
Sketch the graphs of each of the following equations, showing the coordinates of all intercepts. Use a CAS calculator to fi nd the coordinates of
the turning points, rounding to 2 decimal places as appropriate.
a y x x== 2( −−1)(x++2) b y==−−(x++3) (2 x−−1)2
ThINk WrITe/drAW
a 1 State the function. a y = x2(x − 1)(x + 2)
2 Find the y-intercept. When x = 0, y = 0
The y-intercept is 0.
eBookplus eBookplus
Tutorial
Worked example 28
On a Calculator page, defi ne the polynomial P(x) by completing the entry line as:
Define p x( )=x4−x3−7x2+5x+10 Then press ENTER ·.
To fi nd the y-intercept, complete the entry line as: p(0)
Then press ENTER ·.
To fi nd the x-intercepts, complete the entry line as: solve ( ( )p x = 0, )x
Then press ENTER ·.
To sketch the graph of P(x), open a Graphs page.
Complete the function entry line as:
f x1( )= p x( )
Then press ENTER ·.
int-0521
3 Find the x-intercepts. When y = 0,
0 = x2(x − 1)(x + 2) x = −2, 0, 1
4 State the x-intercepts, noting where the graph touches and where it cuts the x-axis.
The graph touches the x-axis at x = 0.
The other x-intercepts are −2 and 1.
5 State the coordinates of the turning points.
The minimum turning points are (−1.44, −2.83) and (0.69, −0.40).
The maximum turning point is (0, 0).
6 Sketch the graph of the quartic, using a CAS calculator to assist.
0 x
y
(0, 0) (1, 0) (0.69, −0.40) (−2, 0)
(−1.44, −2.83) b 1 State the function. b y = −(x + 3)2(x − 1)2
2 Find the y-intercept. When x = 0,
y = −(3)2(−1)2
= −9 The y-intercept is −9.
3 Find the x-intercepts. When y = 0,
0 = −(x + 3)2(x − 1)2 x = −3, 1
4 State the points where the graph touches the x-axis from the repeated factors.
The graph touches the x-axis at x = −3 and x = 1.
5 State the coordinates of the turning points.
The maximum turning points are (−3, 0) and (1, 0), and the minimum turning point is (−1, −16).
6 Sketch the graph of the quartic, using a graphics calculator to assist.
0 x
y
(−1, −16) (0, −9)
(1, 0) (−3, 0)
Worked exAMple 29
Determine the equation of the graph shown.
ThINk WrITe/dISplAy
Method 1: Technology-free
1 State the x-intercepts. The x-intercepts are −3, −1, 1, 2.
2 Write the equation using factor form with a
dilation factor of a. y = a(x + 3)(x + 1)(x − 1)(x − 2)
3 State the y-intercept. The y-intercept is 3.
4 Substitute the coordinates of the point where the graph crosses the y-axis into the equation.
(0, 3) ⇒ 3 = a(0 + 3)(0 + 1)(0 − 1)(0 − 2)
5 Solve the equation to find a. 3 = a × 6
a=12
6 Write the equation. y=12(x−1)(x−2)(x+3)(x+1) Method 2: Technology-enabled
1
2 Write the equation. y x x x x
=( −2)( −1)( +1)( +3) 2
x y
−3 0 1
3
−1 2
On a Calculator page, complete the entry line as:a× + × + × − × − →(x 3) (x 1) (x 1) (x 2) y Then press ENTER ·.
To calculate the value of a, complete the entry line as:
solve(y=3, ) |a x=0 Then press ENTER ·.
To find the equation of y, complete the entry line as:
y
|
a = 1Then press ENTER ·.2
Worked exAMple 30
Sketch the graph of y== −−x4−−2x2,∈∈x( , ]−−1 1 , using the unrestricted function as a guide. State the domain and the range in each case.
ThINk WrITe/drAW
1 State the function. y = −x4 − 2x2, x ∈ (−1, 1]
2 Find the y-intercept. When x = 0,
y = −(0)4 − 2(0)2
= 0
3 State the y-intercept. The y-intercept is 0.
4 Find the x-intercepts. When y = 0
−x4 − 2x4= 0
5 Factorise the quartic expression. −x2(x2+ 2) = 0
6 Solve for x. x = 0 is the only solution
(as x2 + 2 ≠ 0).
7 State the x-intercepts. The only x-intercept is 0.
8 Find y when x is one end point of the domain. When x = −1,
y = −(−1)4 − 2(−1)2
= −3
9 State the coordinates and whether it is an open or closed point.
(−1, −3) is an open end point.
10 Find y when x is the other end point of the domain. When x = 1,
y = −(1)4 − 2(1)2
= −3
11 State the coordinates and whether it is an open or closed point.
(1, −3) is a closed end point.
12 Sketch the graph of the quartic, using knowledge of basic shapes or a CAS calculator to assist, over the domain.
y
0 x
y = −x4− 2x2 (0, 0)
(−1, −3) (1, −3)
13 State the domain, which is given with the rule.
The domain is (−1, 1].
14 From the graph, state the range. The range is [−3, 0].
Quartic graphs General equation is
1. y = ax4 + bx3 + cx2 + dx + e.
Basic shapes of quartic graphs:
2.
(a) If a > 0:
0 y
x
y = ax4
0 y
x
y = ax4+ cx2, c ≥ 0
b 0 c
y
x
y = ax2(x − b)(x − c)
0
b c
y
x
y = a(x − b)2(x − c)2
b 0 c
y
x
y = a(x − b)(x − c)3
b c 0 d e
y
x
y = a(x − b)(x − c)(x − d)(x − e)
(b) If a < 0, then the reflection through the x-axis of the types of graph in the figures above is obtained.
reMeMBer
Quartic graphs
1 We 27 Sketch the graph of each of the following, showing all intercepts.
a y = (x − 2)(x + 3)(x − 4)(x + 1) b y = 2x4 + 6x3 − 16x2 − 24x + 32 c y = x4 − 4x2 + 4
d y = 30x − 37x2 + 15x3 − 2x4 e y = 6x4 + 11x3 − 37x2 − 36x + 36
2 We 28 Sketch the graph of each of the following equations, showing the coordinates of all intercepts. use a calculator to find the coordinates of the turning points, rounding to 2 decimal places as appropriate.
a y = x2 (x − 2)(x − 3) b y = −(x + 1)2 (x − 1)2 c y = (x − 1)2(x + 1)(x + 3) d y = (x + 2)3 (1 − x)
3 MC Consider the function f (x) = x4 − 8x2 + 16.
a When factorised, f (x) is equal to:
A (x + 2)(x − 2)(x − 1)(x + 4) B (x − 1)(x − 4)(x + 4) C (x + 3)(x − 2)(x − 1)(x + 1) D (x − 2)3 (x + 2) E (x − 2)2 (x + 2)2
b The graph of f (x) is best represented by:
A y
0 x
−2 2
−16
B y
0 x
–2 2
16
C y
0 x 16
−2 2
D y
0 x 16
−2 2
E y
0 x 4
−2 2
c If the domain of f (x) is restricted to [−2, 2], then the range is:
A [0, 16] B [0, 10] C [−2, 12]
D R+ E [0, ∞)
d If the range of f (x) is restricted to (0, 25) then the maximal domain is:
A [−2, 3) B (−2, 3) C (−3, 2) D (−3, 3) E (−3, 4) e If the domain of f (x) is restricted to (−1, 0), then the range is:
A (0, 16) B (0, 4) C (−1, 9) D (9, 16) E [9, ∞) f If the domain of f (x) is restricted to [0, ∞), then the range is:
A R B R+ C [0, ∞) D [0, 16) E [2, ∞)
exerCISe
1G
eBookplus eBookplus Digital doc SkillSHEET 1.8 Solving quartic
equations
eBookplus eBookplus Digital doc Spreadsheet 105 Quartic graphs
— factor form
4 We 29 Determine the equation of each of the following graphs.
a
x y
0 6
−2 −1 1 3
b
x y
0 2 4
−1 8
5 We 30 Sketch the graph of each of the following restricted functions, using the unrestricted function as a guide. State i the domain and ii the range in each case.
a y = (2 − x)(x2 − 4)(x + 3), x ∈ [2, 3] b y = 9x4 − 30x3 + 13x2 + 20x + 4, x ∈ (−2,−1]
c y = −(x − 2)2(x + 1)2, x ∈ (−∞, −2] d y = 4x2 − x4, x ∈ [−3, −2]
6 The function f (x) = x4+ ax3− 4x2+ bx + 6 has x-intercepts (2, 0) and (−3, 0). Find the values of a and b.
7 The function f (x) = x4+ ax3+ bx2− x + 6 has x-intercepts (1, 0) and (−3, 0). Find the values of a and b.
8 The functions y = (a − 2b)x4− 3x − 2 and y = x4− x3+ (a + 5b)x2− 5x + 7 both have an x-intercept of 1. Find the value of a and b.
eBookplus eBookplus Digital doc
Investigation Quartics and
beyond
SuMMAry
Pascal’s triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Binomial theorem
(ax b) n ( ) ( ) ..
ax b n
ax b
n n n
+ =
+
− +
0 0 1 1 ..+ ( ) ( )
−
+
−
n
n ax b n
n ax b
n n
1 1 0
Notes
Indices add to
1. n.
There are
2. n + 1 terms in the expansion.
The (
3. r + 1)th term is n
r ax n r rb
( ) − . Polynomials
If
• P(x) = an x n+ an − 1 x n − 1+ . . . + a2 x2+ a1 x + a0 and n is a non-negative integer then P(x) is a polynomial of degree n and an, an − 1, . . . , a2, a1 are called coefficients and ∈ R.
Remainder theorem:
•
If P(x) is divided by (x − a), then the remainder is P(a). If P(x) is divided by (ax + b) then the remainder is P −
b a . Factor theorem:
• If
1. P(a) = 0, then (x − a) is a factor of P(x) or if (ax + b) is a factor of P(x), then P b
a
−
= 0 If (
2. x − a) is a factor of P(x) then a must be a factor of the term independent of x.
Linear graphs
Linear equations are polynomials of degree 1.
•
General equation is
• ax + by + c = 0
or y = mx + c
where m = gradient c = y-intercept The gradient
• m y y
x x
= −
2− 1
2 1
Equation if a point and the gradient is known:
• y − y1 = m(x − x1)
Parallel lines have the same gradient.
• If
• m1 and m2 are the gradients of perpendicular lines, then:
m1× m2= −1 or m
1 m
2
= −1
Quadratic graphs
Quadratic equations are polynomials of degree 2.
•
General equation is
• y = ax2 + bx + c
Quadratic formula is
• x b b ac
= − ± a2 −4 2 Discriminant
• = b2 − 4ac and
if
1. b2− 4ac > 0, there are 2 x-intercepts (and if b2− 4ac is a perfect square, the intercepts are rational) if
2. b2− 4ac = 0, there is 1 x-intercept if
3. b2− 4ac < 0, there are no x-intercepts.
The power form or turning point form of the quadratic is:
•
y = a(x − b)2+ c and the turning point is (b, c).
The equation of the axis of symmetry and the
• x-value of the turning point of a parabola is −b
a 2 . The axis of symmetry is halfway between the
• x-intercepts.
Cubic graphs
Cubic equations are polynomials of degree 3.
•
General equation is
• y = ax3+ bx2+ cx + d
Basic shapes of cubic graphs:
• If 1. a > 0:
Positive cubic y
x
Basic form
y y = a(x − b)3 + c
(b, c) x
Factor form
y = a(x − b)(x − c)(x − d) where a > 0
y
c d x b
Repeated factor y
a b x
y = (x − a)2 (x − b)
If
2. a < 0, then the reflections through the x-axis of the types of graph in the above figures are obtained.
Quartic graphs
Quartic equations are polynomials of degree 4.
•
General equation is
• y = ax4+ bx3+ cx2+ dx + e
Basic shapes of quartic graphs:
• If 1. a > 0:
y
0 x y = ax4
y
0 x
b c
y = a(x − b)2(x − c)2
y
0 x y = ax4+ cx2, c ≥ 0
y
0 x
b c
y = a(x − b)(x − c)3 y
0 x
b c
y = ax2(x − b)(x − c)
y
0 x
b c d e
y = a(x − b)(x − c)(x − d)(x − e) 2. a < 0, then reflection through the x-axis of the types of graph above is obtained.If
Note: It is possible to translate the cubic and quartic graphs shown in the cubic graphs and quartic graphs sections above.
Functions
A function is fully defined if the rule and domain are given.
•
The domain of a function is the set of values of
• x for which the function is defined.
The range of a function is the set of values of
• y for which the function is fully defined.
Restricted domains can be represented by interval notation:
•
[a, b] = {x: a ≤ x ≤ b} (a, b) = {x: a < x < b} [a, b) = {x: a ≤ x < b}