Master cubic and quartic functions the easy way, with plain English intuition, factor form, turning points, intercepts and an auto marked practice test. VCE Maths Methods Units 3 and 4.
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Stretch a piece of string into a wave with two humps, or one gentle bend, and you have just drawn a polynomial. These are the smooth, friendly curves of Maths Methods: no breaks, no sharp corners, no holes. A cubic has up to one bend each way, a quartic can wiggle a little more, and once you can read their shape straight from the way they are written, sketching them stops being guesswork and becomes a quick, reliable recipe.
What a polynomial really is
A polynomial is just a sum of powers of x with number coefficients, like y=2x3−4x2+x−5. The degree is the highest power, and it tells you almost everything about the overall shape.
A cubic has degree 3. Its graph can have at most two turning points and crosses the x axis up to three times.
A quartic has degree 4. It can have up to three turning points and up to four x intercepts.
The single most useful trick is that the curve is completely smooth. There are no jumps and no corners, so once you know where it meets the axis and which way the ends point, the rest of the sketch joins up naturally.
Factor form is the shortcut
The friendliest way to write a polynomial for sketching is factor form, a product of brackets like
y=a(x−p)(x−q)(x−r).
Each bracket switches off, equalling zero, at one value of x, and that is exactly where the curve meets the x axis. So you can read the intercepts straight off:
x=p,x=q,x=r.
Watch the signs. The bracket (x+2) is zero when x=−2, not x=2. Flipping every sign by habit is one of the most common slips in the exam. For the y intercept, set x=0 and multiply the brackets out.
Repeated factors bend the curve
What happens at an intercept depends on how many times its factor appears.
A single factor, like (x−1), means the curve passes straight through the axis.
A double factor, a squared bracket like (x−2)2, means the curve touches and turns. That intercept is a turning point sitting on the axis.
A triple factor, a cubed bracket like (x−2)3, gives a stationary point of inflection: the curve flattens to zero gradient and then crosses through.
For y = (x+3)(x-2)^2 the single factor makes the curve cross straight through at x = -3, while the squared factor makes it touch the axis and turn at x = 2.
This is a favourite of examiners, and the report comment that keeps coming back is that students give only the x value when a full coordinate was asked for, or confuse a touch with a crossing. When a double factor produces a turning point, write the answer as a coordinate such as (2,0), not just x=2.
The leading coefficient sets the ends
The leading term, the one with the highest power, decides which way the far ends of the graph point. For a cubicy=ax3+…:
If a>0 the curve falls on the far left and rises on the far right.
If a<0 it does the opposite, rising on the left and falling on the right.
For a quarticy=ax4+…, both ends point the same way: both up when a>0, both down when a<0, like a wide letter U or an upside down one. Forgetting the minus sign on a and drawing the wrong way up is a frequent and costly error.
How to sketch one, step by step
The shape work is the hard part. The sketch itself is a short routine you repeat every time.
Factorise if you can, then read the x intercepts from the brackets, watching the signs.
At each intercept decide cross, touch and turn, or flatten, from whether the factor is single, double, or triple.
Find the y intercept by setting x=0.
Use the sign of the leading coefficient to point the two far ends, then join everything with a single smooth curve.
See this recipe in action in the Worked Examples tab, then test yourself in Try It.
Lock it in with active recall
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
In factor form y=a(x−p)(x−q)(x−r), where are the x intercepts?
At x=p, x=q and x=r — set each bracket to zero. Watch the signs: (x+2) is zero at x=−2.
What does a single, double and triple factor each do at the axis?
Single: the curve crosses straight through. Double (squared): it touches and turns. Triple (cubed): a stationary point of inflection — flattens, then crosses.
How do you find the y intercept of a polynomial?
Set x=0 and evaluate, i.e. the y intercept is f(0).
For a cubicy=ax3+…, how do the ends behave when a>0 versus a<0?
If a>0 the curve falls on the far left and rises on the far right; if a<0 it rises on the left and falls on the right.
How does the degree decide which way the two ends point?
Odd degree: the ends point in opposite directions. Even degree (like a quartic): both ends point the same way.
Why must you write a turning point on the axis as a full coordinate, not just an x value?
Examiner reports flag answers giving only the x value. A touch-and-turn at x=2 must be written (2,0).
Recall · Transformations of Graphs
The graph y=(x+4)2−7 comes from y=x2 by what translation?
4 units left and 7 units down. The +4 inside acts on x (opposite of the sign), the −7 outside lowers y directly.
Recall · Exponential Functions
How do you solve e2x−5ex+4=0?
Let a=ex to get the quadratic a2−5a+4=0, factorise, then solve ex=a for each positive root (reject any a≤0).
Worked examples
Worked Example 1Counting x-intercepts as a parameter varies, from a real exam
For f(x)=(x+1)(x+a)(x−2)(x−2a), find the values of a for which the graph of y=f(x) has (i) exactly three x-intercepts; (ii) exactly four x-intercepts.
1
The four roots are −1,−a,2,2a. Exactly three intercepts means exactly one coincidence occurs between roots.
−a=2⇒a=−2;2a=−1⇒a=−21;−a=2a⇒a=0
2
For exactly four intercepts all four roots are distinct. Note a=1 gives only two distinct roots (−1 and 2 both doubled), so it must be excluded.
a∈R∖{−2,−21,0,1}
Answer
(i)a=−2,a=−21,a=0;(ii)a∈R∖{−2,−21,0,1}
VCAA 2024 Mathematical Methods Exam 2, Section B Q1b
Worked Example 2Sketching a cubic-quartic, from a real exam
Sketch the graph of y=g(x) where g(x)=4x3−3x4, labelling the stationary points and axial intercepts with their coordinates.
1
Find the axial intercepts by solving g(x)=0 in factored form.
x3(4−3x)=0⇒x=0orx=34
2
Mark the stationary point of inflection at (0,0) and the local maximum at (1,1); the curve falls to −∞ on either side.
SPI (0,0),max (1,1),x-ints (0,0),(34,0)
Answer
Cubic-quartic with SPI at (0,0),local max at (1,1),x-intercepts (0,0) and (34,0)
VCAA 2025 Mathematical Methods Exam 2, Section B Q1b
Worked Example 3Sketching a cubic from its factors
Sketch the graph of y=(x+2)(x−1)(x−3), showing all axis intercepts as coordinates.
1
Each bracket equals zero at one x value. Set each factor to zero to read the x intercepts straight off.
x=−2,x=1,x=3
2
Find the y intercept by putting x=0 into the rule.
y=(2)(−1)(−3)=6
3
The leading term is positive x3, so the curve rises from bottom left to top right and crosses the axis cleanly at each of the three single factors.
intercepts (−2,0),(1,0),(3,0),(0,6)
Answer
A positive cubic crossing at (−2,0),(1,0),(3,0) with y intercept (0,6)
Worked Example 4A repeated factor changes the shape
Describe the behaviour of y=(x+1)(x−2)2 at each of its x intercepts.
1
Set each factor to zero. The factor (x−2) is squared, so x=2 is a repeated root.
x=−1(single),x=2(double)
2
A single factor means the curve passes straight through the axis. A squared (double) factor means the curve just touches the axis and turns back, so (2,0) is a turning point.
at x=−1 cross;at x=2 touch and turn
3
Check the y intercept by setting x=0.
y=(1)(−2)2=4
Answer
Crosses at (−1,0), touches and turns at (2,0),y intercept (0,4)
Worked Example 5Effect of the leading coefficient
The cubic y=a(x−1)(x+2)(x−4) passes through the point (0,16). Find a and state the end behaviour.
1
Substitute the known point (0,16) to make an equation in a.
16=a(0−1)(0+2)(0−4)
2
Evaluate the brackets, then solve for a.
16=a(−1)(2)(−4)=8a⇒a=2
3
Since a=2>0, the leading term behaves like 2x3. A positive cubic falls on the far left and rises on the far right.
a=2>0⇒down on the left, up on the right
Answer
a=2,positive cubic: as x→−∞,y→−∞ and as x→+∞,y→+∞
Practice questions
Practice test
Try it yourself
9 questions, 11 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.For the parabola with equation y=ax2+2bx+c, where a,b,c∈R, the equation of the axis of symmetry is:
1mark
Show worked solution
The linear coefficient here is 2b, so the axis of symmetry x=−2AB=−2a2b=−ab. The common trap is to use b in place of the actual coefficient 2b and answer −2ab.
VCAA 2023 Mathematical Methods Exam 2, Section A Q2
Q2.Let f:R→R, f(x)=x(x−2)(x+1). State the coordinates of all axial intercepts of f.
1mark
Work this on paper. The worked solution appears once you submit.
Show worked solution
Setting each factor to zero gives intercepts at x=0,2,−1, all with y=0, and the y-intercept coincides with (0,0). The axial intercepts are (0,0), (2,0) and (−1,0). The report notes coordinates (not just x-values) were required.
VCAA 2023 Mathematical Methods Exam 2, Section B Q1a
Q3.Let f(x)=(x+2)2(x−5) and g(x)=x+2. Write down the values of x for which f(x)g(x)≥0.
1mark
Work this on paper. The worked solution appears once you submit.
Show worked solution
The product is (x+2)3(x−5), with roots x=−2 (triple) and x=5. This is negative only on the interval (−2,5), so the product is ≥0 for x≤−2 or x≥5 (including equality at the roots). The report notes this was poorly answered, with common wrong answers including x≤2 (sign error) and −2≤x≤5.
VCAA 2025 Mathematical Methods Exam 1, Q7d.ii
Q4.The x intercepts of y=(2x−1)(x+3)(x−5) are:
1mark
Need a hint?
Set each bracket to zero and solve. For 2x−1, do not forget to divide by the coefficient 2.
Show worked solution
Set each factor to zero. From 2x−1=0 we get x=21, from x+3=0 we get x=−3, and from x−5=0 we get x=5. Option C is the common sign slip: students flip the sign of every root, forgetting that x+3=0 gives x=−3, not x=3. Option D ignores the coefficient 2 in the first bracket.
Q5.The graph of y=(x−3)2(x+1) meets the x axis where it:
1mark
Need a hint?
Count how many times each factor appears: a squared bracket behaves differently from a single one.
Show worked solution
A squared factor produces a turning point on the axis, so the double factor (x−3)2 makes the curve touch and turn at x=3. The single factor (x+1) makes the curve cross straight through at x=−1. Option C swaps the two behaviours, the most common error: the repeated factor, not the single one, is where the curve turns back on itself.
Q6.As x→−∞ and as x→+∞, the graph of y=−2x3+5x−1 behaves like:
1mark
Need a hint?
Only the leading term matters for the ends. Check the sign in front of the highest power.
Show worked solution
Only the leading term decides the end behaviour. Here it is −2x3, a negative cubic, so the curve comes down from top left and falls to bottom right: as x→−∞, y→+∞, and as x→+∞, y→−∞. Option B is the positive cubic shape and is the usual slip when the minus sign on the leading coefficient is overlooked. Options D and A describe quartic (even degree) end behaviour, where both ends point the same way.
Q7.A cubic of the form y=a(x+2)(x−1)(x−4) passes through (0,−16). The value of a is:
1mark
Need a hint?
Substitute the point into the rule, evaluate the brackets carefully with their signs, then solve for a.
Show worked solution
Substitute (0,−16): −16=a(2)(−1)(−4)=8a, so a=−2. Option A drops the negative sign from the y value. Option D reads a straight off the y intercept without dividing by the product of the brackets, and option C forgets to solve for a at all.
Q8.The quartic y=(x−2)3(x+1) has, at x=2, a:
1mark
Need a hint?
Think about what a factor to the power three does at the axis, compared with a squared factor.
Show worked solution
A factor raised to the power three, a triple factor, gives a stationary point of inflection on the axis: the curve flattens to zero gradient and crosses through. Option B describes a double (squared) factor, where the curve touches and turns but does not cross. The triple factor crosses, so it is not a touch and turn. Polynomials are smooth everywhere, so a sharp corner never occurs.
Q9.For y=(x+3)(x−2)2, state the coordinates of all axis intercepts and describe the shape of the graph at each x intercept.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Set each factor to zero. The single factor gives x=−3 and the squared factor gives a repeated root at x=2.
The x intercepts are (−3,0) and (2,0).
For the y intercept, put x=0:
y=(3)(−2)2=3×4=12,
so the y intercept is (0,12).
At x=−3 the single factor means the curve crosses the axis. At x=2 the squared factor means the curve touches and turns, so (2,0) is a turning point. The leading term is positive x3, so the curve rises from bottom left to top right overall. Full coordinates are required, not just the x values.
Frequently asked questions
How do you find the x intercepts of a polynomial in factor form?
Set each bracket equal to zero and solve. Watch the signs: the bracket (x + 2) is zero when x equals negative 2, not positive 2. Flipping the sign by habit is one of the most common exam slips.
What is the difference between a curve crossing, touching, and flattening at an x intercept?
It depends on how many times the factor appears. A single factor means the curve crosses straight through the axis, a double (squared) factor means it touches and turns, and a triple (cubed) factor gives a stationary point of inflection where it flattens then crosses.
How does the leading coefficient affect the shape of a cubic or quartic graph?
The leading term decides which way the far ends point. For a cubic, a positive coefficient falls on the left and rises on the right, and a negative one does the opposite. For a quartic, both ends point the same way, both up when positive and both down when negative.
