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# Secondary Magazine - Issue 120: It Stands to Reason

Created on 10 March 2015 by ncetm_administrator
Updated on 29 April 2015 by ncetm_administrator

# It Stands to Reason In Praise of Curve Sketching

Why is “pencil and paper” curve sketching still an important activity in mathematics when we have such amazing graph plotting software available in the 21st century such as Autograph and Geogebra? The answer lies in the process that learners go through in order to provide information for the sketch graph, and the mathematical reasoning that underpins this process and facilitates the establishment of the key features of the sketch graph. That’s why a question along the lines of “How many roots does the equation $\inline \dpi{80} \fn_jvn x^{2} = 2^{x}$ have?” has been such a stalwart of university entrance interviews for so many years.

This process of reasoning and discovery, for example answering questions such as

• “Does the curve cross either or both of the coordinate axes, and if so, how many times and where?”
• “Does it have any turning points, and if so, how many, what type and where?”
• “Are there any asymptotes and/or what happens for large values of $\inline \dpi{80} \fn_jvn x$?”

can take place before or after the actual features of the graph are suggested via graph plotting software, and the two can even be used hand in hand. Here is an example of some student thinking, working and reasoning with teacher comments, and prompts for further reasoning, in italics. It’s a complicated example, but one which demonstrates the full power of curve sketching because it entails so many steps. Some simpler examples are suggested at the end of the article.

Problem. Investigate the graph of $\inline \dpi{80} \fn_jvn y = 6ln(x^{2}+1) - x$

Student 1

Thinking. Start by just drawing the graph in Geogebra.

Teacher

To get a feel for an unfamiliar graph, and then start to investigate the features more fully by asking questions like those above. In answering these questions, mathematical reasoning will come to the fore.

Student 2

Teacher

Some colleagues will no doubt balk at this initial use of software and insist on investigation and reasoning first in a curve sketching activity. However, the students can discuss straightaway the benefits but also the limitations of this computer-generated graph: a key one is that we only see the shape of the graph over a very limited domain.

Student 1

Thinking. It looks like the graph crosses the coordinate axes near the origin but where and how many times?

Student 2

Working. When $\inline \dpi{80} \fn_jvn x = 0$ (i.e. where the graph crosses the y-axis), $\inline \dpi{80} \fn_jvn y=6ln1-0=0$.
Reasoning. This tells me that the graph crosses the y-axis at only one point and this is at (0,0). There are no other intercepts with the y-axis.

Student 2

Working. When $\inline \dpi{80} \fn_jvn y = 0$ (i.e. where the graph crosses the x-axis), $\inline \dpi{80} \fn_jvn 6ln(x^{2}+1) - x = 0$.
Thinking. This is a much trickier equation to solve or indeed even to be able to say how many roots it has.

Student 1

Reasoning. One of the solutions is $\inline \dpi{80} \fn_jvn x = 0$as we already know that the graph passes through (0,0). So the graph crosses the x-axis at least once.
Thinking. Are there any more intercepts with the x-axis? There might be but $\inline \dpi{80} \fn_jvn 6ln(x^{2}+1) - x = 0$ is an equation that will need an approximate method of solution using an accurately plotted graph (which we are trying to avoid at the moment) or numerical methods.
Working. Writing the equation as $\inline \dpi{80} \fn_jvn x = 6ln(x^{2}+1)$ and using an iterative method along with a calculator or spreadsheet or graph plotting software leads to roots x = 0 or x = 45.93 (4 s.f.).
Reasoning. There are at least two intercepts with the x-axis at (0,0) and (45.93,0).

Student 2

Thinking. The root $\inline \dpi{80} \fn_jvn x = 45.93$ is a surprise! I can see two intercepts with the x-axis at present but neither of them is 45.93. What’s the value of the root between 0 and 1? Are there any more?

Student 1

Working. Rewriting the equation as $\inline \dpi{80} \fn_jvn x = \sqrt{e^{x/6} - 1}$ (Further Mathematics students need to be able to explain why different iterative functions converge to different roots) and using an iterative method along with a calculator or spreadsheet or graph plotting software leads to the root $\inline \dpi{80} \fn_jvn x = 0.1690$ (4 s.f.).
Reasoning. I have found three intercepts with the x-axis at (0,0), (0.1690,0) and (45.93,0).

Student 2

Thinking. How can I be sure that there are no more intercepts? After all, I wasn’t expecting 45.93.

Student 1

Thinking. Are there any turning points, and if so can I deduce anything from them?
Working $\inline \dpi{80} \fn_jvn \frac{dy}{dx} = \frac{12x}{\left ( x^{2} \right + 1)} - 1 = - \frac{\left ( x^{2} - 12x+1 \right )}{\left ( x^{2} +1\right )}$ and so on for turning points $\inline \dpi{80} \fn_jvn x^{2} - 12x + 1 = 0$ and so $\inline \dpi{80} \fn_jvn x=6\pm \sqrt{35}$.
Reasoning. This tells me that there are exactly two turning points on the graph. One of them is where $\inline \dpi{80} \fn_jvn x = 6 - \sqrt{35}$ which is small (close to zero) but positive, and $\inline \dpi{80} \fn_jvn y = 6ln((6 - \sqrt{35})^{2}+ 1) - (6 - \sqrt{35})< 0$ (explain why) and small (explain why) so this turning point is just below the x-axis. The second is where $\inline \dpi{80} \fn_jvn x = 6 + \sqrt{35}$ which is much larger, positive and close to but less than 12, and $\inline \dpi{80} \fn_jvn y \approx 6ln145 - 12 > 0$ (explain why using approximations for powers of $\inline \dpi{80} \fn_jvn e$ like $\inline \dpi{80} \fn_jvn e^{3}\approx 20$) so this turning point is well above (explain why) the x-axis.

Student 2

Thinking. What type of turning points are they?
Working. Let’s look at another, more familiar, sketch graph of $\dpi{80} \fn_jvn y = x^{2}- 12x + 1$ in order to ascertain the nature of $\dpi{80} \fn_jvn \frac{dy}{dx}$ for the original function.

Reasoning. For $\dpi{80} \fn_jvn x< 6-\sqrt{35}$$\dpi{80} \fn_jvn x^{2}- 12x + 1> 0$ and so the original $\dpi{80} \fn_jvn \frac{dy}{dx}< 0$ ie negative.
For $\dpi{80} \fn_jvn 6-\sqrt{35} < x< 6+\sqrt{35}$$\dpi{80} \fn_jvn x^{2}-12x+1< 0$ and so the orignal $\dpi{80} \fn_jvn \frac{dy}{dx}> 0$ie positive.

This tells me that the turning point at $\dpi{80} \fn_jvn x=6+\sqrt{35}$ is a minimum point as the gradient of the graph changes from negative to positive through zero at this point.
For $\inline \dpi{80} \fn_jvn x> 6+\sqrt{35}$, $\inline \dpi{80} \fn_jvn x^{2}-12x+1> 0$ and so the original $\dpi{80} \fn_jvn \frac{dy}{dx}< 0$ i.e. negative.

This tells me that the turning point at $\dpi{80} \fn_jvn x=6+\sqrt{35}$ is a maximum point as the gradient of the graph changes from positive to negative through zero at this point.

I can also deduce from the information about the turning points that there will be an intercept with the x-axis between the two turning points at $\dpi{80} \fn_jvn x=6-\sqrt{35}$ (minimum point) and $\dpi{80} \fn_jvn x=6+\sqrt{35}$ (maximum point), and another intercept with the x-axis beyond the (maximum) turning point at $\dpi{80} \fn_jvn x=6+\sqrt{35}$

Unfortunately I can’t deduce where these intercepts are exactly from this information (I need my numerical methods for this), but I now know that there will be a total of three intercepts with the x-axis and no more.

Student 1

Thinking. Can I get a rough idea where the x-intercepts are? I know one is definitely at $\dpi{80} \fn_jvn x=0$. What about the other two?
Reasoning. The second intercept with the x-axis is approximately twice the distance (assuming approximate local symmetry of the graph)(explain why this is possible) from the origin to $\dpi{80} \fn_jvn (6-\sqrt{35}, 0)$ ie $\dpi{80} \fn_jvn 2(6-\sqrt{35})= 0.1678$ (4 s.f.) compared to the actual value of 0.1690 (4 s.f.).

Student 2

Thinking. What about the third intercept which has a larger value of $\dpi{80} \fn_jvn x$?

Reasoning. For $\dpi{80} \fn_jvn x$ with large modulus, $\dpi{80} \fn_jvn y=6ln (x^{2}+1) - x \rightarrow 12ln\left | x \right |- x$(explain this including why you have introduced the modulus function here) , so our graph will approach the graph of $\dpi{80} \fn_jvn y=12ln\left | x \right |- x$for $\dpi{80} \fn_jvn x$ with large modulus. For positive $\dpi{80} \fn_jvn x$, the equation has a root between 36 and 48 (explain why) which is much closer to 48 than 36 (I previously showed the actual root is $\dpi{80} \fn_jvn x=45.93$ (4 s.f.).

Student 1

Thinking. Finally I can sketch my graph using the intercepts with the axes, the turning points and the behaviour for large, positive values of $\dpi{80} \fn_jvn x$.

The student reasoning here is very good but you could probably improve it or add to it (or ask your students to improve it or add to it). Similarly, the sketch graph is OK but could be improved (how?). The actual graph is below:

“Paper and pencil” curve sketching gives students multiple opportunities to develop their reasoning abilities in gradual stages, from simple graph sketching of basic functions in GCSE Mathematics or at the start of AS Level Mathematics to much more challenging problems such as this example in A Level Mathematics or those found in A Level Further Mathematics. They do so in more than one area of mathematics: they have to link the consequences of algebraic and numerical reasoning to their geometric and graphical understanding of the function whose graph is being sketched. It is this interweaving that makes curve sketching such a rewarding and worthwhile activity. So why not start your next lessons asking your students to sketch and reason about related curves such as

• $\dpi{80} \fn_jvn y=x^{2}-4x+1$
• $\dpi{80} \fn_jvn y=x^{4}-4x^{2}+1$
• $\dpi{80} \fn_jvn y=4^{x}-2^{x+2}+1$
• $\dpi{80} \fn_jvn y=(x^{2}-4x+1)^{2}$

or

• $\dpi{80} \fn_jvn y=\sin^{2}x$
• $\dpi{80} \fn_jvn y=\sin (x^{2})$
• $\dpi{80} \fn_jvn y=1/\sin (x)$
• $\dpi{80} \fn_jvn y=\sin (1/x)$
• $\dpi{80} \fn_jvn y=x\sin (x)$
• $\dpi{80} \fn_jvn y=x\sin (1/x)$

Your students don’t need to be able to differentiate exponential or trigonometric functions (or know the chain or product rules) to investigate these graphs in considerable detail. You could have a “first pass” at sketching them in Y12, and return to them in Y13 for a “second pass”.

Image credit

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