Sketching graphs is a useful skill when you need to check the overall shape of a graph, but don’t need to plot it perfectly. This blog describes sketching graphs of cubic, quartic and reciprocal curves (you should have already learnt sketching graphs of quadratic curves at GCSE) by finding the 𝑥- and 𝑦-intercepts.
You can add more detail to your sketches by including the turning points – this requires using differentiation to find stationary points and is covered in a different blog, here.
To download this blog in PDF or PowerPoint formats, click here. To practice some of the prior-knowledge required for this topic, try this multiple-choice quiz.
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A cubic curve has an equation of order 3 (the highest power of 𝑥 in the equation is 3). In other words, its equation is in the form:
𝑦 = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑
where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers where 𝑎 ≠ 0.
The general shape of cubic curves are shown in the diagrams below. When sketching graphs of cubic functions, the first thing to check is whether 𝑎 is greater than or less than 0 – this defines the orientation of your graph. The shape of that graph will become more accurate with further working.
A quartic curve has an equation with an order of 4; this is of the form:
𝑦 = 𝑎𝑥4 + 𝑏𝑥3 + 𝑐𝑥2 + 𝑑𝑥 + 𝑒
where 𝑎, 𝑏, 𝑐, 𝑑 and 𝑒 are real numbers and 𝑎 ≠ 0.
The general shape of a quartic curve is shown in the diagrams below. Again, when sketching graphs of cubic functions, the value of 𝑎 gives the orientation of the graph:
When sketching graphs of either cubic or quartic functions, first work out the general shape of the curve then calculate the value (or values) of its intercepts. Do this in a similar way to how you would calculate the intercepts for quadratics.
For a curve whose equation is of the form 𝑦 = f(𝑥):
- Solve f(𝑥) = 0 to find the points of intersection with the 𝑥-axis. The values of 𝑥 for which f(𝑥) = 0 are called the roots of the function.
- Calculate 𝑦 = f(0) to get the point of intersection with the 𝑦-axis.
Use all of this information to sketch the curve. Remember, sketching is not the same as plotting the curve – the diagram does not need to be perfectly to scale but it should have the right shape and the values of any intercepts should be clearly labelled.
Example Question 1
Sketch the curve whose equation is 𝑦 = (𝑥 + 1)(𝑥 – 2)(𝑥 – 4).
If you expand the brackets, you’ll see that the leading term would be 𝑥3, so this is a cubic equation, with a positive 𝑥3 term. You therefore know that the graph will have this general shape:
Next, you need to find the intercepts of the curve with the axes.
To find the value of the 𝑥-intercepts, set (𝑥 + 1)(𝑥 – 2)(𝑥 – 4) = 0 and solve for 𝑥.
As the cubic is already factorised, this gives 𝑥 = -1, 𝑥 = 2 and 𝑥 = 4.
To find the value of the 𝑦-intercepts, let 𝑥 = 0:
𝑦 = (0 + 1)(0 – 2)(0 – 4)
𝑦 = 8
You can now sketch the curve. Make sure you mark the values of the intercepts clearly. Remember, the sketch does not need to be to scale but should have the correct shape.
Example Question 2
Sketch the curve with equation 𝑦 = -2𝑥4 + 2𝑥3 + 38𝑥2 – 98𝑥 + 60.
The highest power of 𝑥 is 4, so this is a quartic graph. Since the coefficient of 𝑥4 is negative, the general shape will be one of the following:
To find the values of the 𝑥-intercepts, set -2𝑥4 + 2𝑥3 + 38𝑥2 – 98𝑥 + 60 = 0 and solve for 𝑥. You can use the polynomial solver on your calculator to do this.
-2𝑥4 + 2𝑥3 + 38𝑥2 – 98𝑥 + 60 = 0
𝑥 = 3, 𝑥 = 2, 𝑥 = -5, 𝑥 = 1
To find the y-intercept, let 𝑥 = 0:
𝑦 = -2 × 04 + 2 × 03 + 38 × 02 – 98 × 0 + 60
𝑦 = 60
Sometimes, the solutions to f(𝑥) = 0 might be a little unusual. This can tell you a lot about the graph.
- A repeated root is a root that occurs more than once. If it is repeated an even number of times then the curve will touch, rather than cross, the 𝑥-axis at that point.
- A “non-real” solution means the curve will not intersect the 𝑥-axis at the number of points you might usually expect. If you’re solving on a calculator, this sort of solution might be of the form 𝑎 + 𝑏𝑖, but you can also establish a non-real solution with a negative discriminant.
Example Question 3
Sketch the curve with equation 𝑦 = (𝑥 + 2)2(𝑥2 + 2𝑥 + 5).
If you expanded these brackets, you’d find the highest power of 𝑥 to be 𝑥4 so the curve is quartic. The 𝑥4 coefficient is positive, which tells you something about the shape of the curve.
To find the 𝑥-intercepts, set (𝑥 + 2)2(𝑥2 + 2𝑥 + 5) = 0 and solve for 𝑥. Start by solving (𝑥 + 2)2 = 0. Since the expression 𝑥 + 2 is squared, this has a repeated root of 𝑥 = -2.
There are no real solutions to 𝑥2 + 2𝑥 + 5 = 0, so there are no more 𝑥-intercepts – the curve touches the 𝑥-axis exactly once, at 𝑥 = -2.
Finally, let 𝑥 = 0:
𝑦 = (0 + 2)2(02 + 2 × 0 + 5)
𝑦 = 20
The sketch will look like this:
Notice that, at this stage, you do not always have a way to work out the exact turning points of the curves or whether the curves actually have the number of turning points you might expect. In fact, as a quartic the curve above could look more like the curve of quadratic, with a single minimum at -2 and 𝑦-intercept at 20.
To work out the turning points, you need to use calculus to differentiate your function and find the number of stationary points. This isn’t covered in this blog, but you can read more about it here.
A reciprocal curve has some special features which make it easily recognisable. The general equation of a reciprocal curve is \(y=\frac{a}{\text{f}(x)}\) where 𝑎 is a non-zero constant and f(𝑥) is a polynomial function. At this stage, we are only interested in functions such as \(y=\frac{2}{x}\) or \(y=\frac{1}{x^2}\).
Take the function \(y=\frac{1}{x}\). If you chose to plot (instead of sketch) the graph of this function, you might begin with a table of values:
As the absolute value of 𝑥 gets larger, the value of 𝑦 gets closer to zero. The graph will therefore approach zero as 𝑥 approaches positive, or negative, infinity.
You can generalise this result for graphs of reciprocals of the form \(y=\frac{a}{x}\) or \(y=\frac{a}{x^2}\):
If 𝑎 < 0, the curves will be reflected in the 𝑥-axis but will otherwise have the same shape.
When sketching graphs of reciprocal functions, you don’t need to work out the intercepts. Look at the graphs above – notice how the curves of both \(y=\frac{a}{x}\) and \(y=\frac{a}{x^2}\) have a horizontal asymptote at 𝑦 = 0 (the 𝑥-axis) and a vertical asymptote at 𝑥 = 0 (the 𝑦-axis). This is because setting either 𝑥 = 0 or 𝑦 = 0 in the equations of these curves will give a result that is undefined. The graphs will approach the axes but never reach them.
Practice Questions
1. In each case, sketch the curve and clearly indicate any points of intersection with the axes:
a. 𝑦 = (𝑥 + 2)(𝑥 + 3)(𝑥 – 1)
b. 𝑦 = (𝑥 – 4)(𝑥 – 1)(𝑥 + 5)(𝑥 + 7)
c. 𝑦 = 𝑥(𝑥 + 3)2
d. 𝑦 = (𝑥 + 2)2(𝑥 – 1)2
e. \(y=\frac{2}{x}\)
f. \(y=-\frac{1}{x^2}\)
2. The diagram shows the graph of a function 𝑦 = 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 𝑐, where 𝑎, 𝑏 and 𝑐 are real constants. Find the values of 𝑎, 𝑏 and 𝑐.
3. The diagram shows the graph of the function 𝑦 = (-𝑥 + 𝑎)(𝑥 + 𝑏)(𝑥 + 𝑐), where 𝑎, 𝑏 and 𝑐 are real constants such that 𝑎 < 𝑏 < 𝑐. Write down the values of 𝑎, 𝑏 and 𝑐 and state the coordinates of the 𝑦-intercept.
4. The diagram shows the graph of a function 𝑦 = 𝑎𝑥4 + 𝑏𝑥3 + 𝑐𝑥2 + 𝑑𝑥 + 𝑒, where 𝑎, 𝑏, 𝑐, 𝑑 and 𝑒 are real constants. Work out the equation of the function.
Answers
