Two-way functions
This gives you an opportunity to explore roots and asymptotes of functions, both by identifying properties that functions have in common and also by trying to find functions that have particular properties. You may like to use the list of functions in the Hint, which includes enough functions to complete the table plus some extras.You might like to work on this problem in a pair or small group,
or to compare your table to someone else's to see where you have used the same functions and where not.
Problem
Each column and row heading in the following table is a property that a function may or may not have. A function can appear in a cell if it has the properties in the corresponding row and column.
We have omitted some headings, and some entries in cells. Can you complete the table?
You might find it helpful to draw some sketches. You could use graph-sketching software such as Desmos to help you, but try to do the sketching by hand first, before reaching for a computer or calculator!
Make sure that you can explain why each function has the desired properties.
| $y$-axis is an asymptote | passes through origin | |||
|---|---|---|---|---|
| $x = 1$ is a root | $y = \vert x-1\vert$ | $y = -3x+3$ | ||
| has exactly two roots | $y = x(x-2)$ | |||
| $y = \dfrac{1}{x}$ for $x \neq 0$ | $y = \dfrac{1}{(x-1)^2}$ for $x \neq 1$ | $y = \dfrac{3}{x+1}$ for $x \neq -1$ | ||
| $y \to \infty$ as $x \to \infty$ | $y = 2+(x-1)^4$ |
Can you complete the table using a different function in every cell? By contrast, how few different functions can you use in the table?
Did you have any choice about the column and row headings?
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Getting Started
Here are some functions that might (or might not) prove helpful when trying to complete the table. You could sketch each to see whether it has properties that you find useful.
$y = (x-1)^2$
$y = x(1-x)$
$y = \dfrac{1}{\vert x \vert}$ for $x \neq 0$
$y = 1 + \dfrac{1}{x}$ for $x \neq 0$
$y = (x-1)^2 (x+3)$
$y = \dfrac{x-1}{x}$ for $x \neq 0$
$y = x^3$
$y = \dfrac{x}{(x-1)^2}$ for $x \neq 1$
$y = -1 + \dfrac{1}{x^2}$ for $x \neq 0$
$y = \vert x \vert$
$y = x(x+7)$
$y = \dfrac{3}{x^2}$ for $x \neq 0$
$y = x + \dfrac{1}{x}$ for $x \neq 0$
$y = x^2 + 3$
$y = x$
$y = \dfrac{1}{x^3}$ for $x \neq 0$
$y = (x-1)^2$
$y = x(1-x)$
$y = \dfrac{1}{\vert x \vert}$ for $x \neq 0$
$y = 1 + \dfrac{1}{x}$ for $x \neq 0$
$y = (x-1)^2 (x+3)$
$y = \dfrac{x-1}{x}$ for $x \neq 0$
$y = x^3$
$y = \dfrac{x}{(x-1)^2}$ for $x \neq 1$
$y = -1 + \dfrac{1}{x^2}$ for $x \neq 0$
$y = \vert x \vert$
$y = x(x+7)$
$y = \dfrac{3}{x^2}$ for $x \neq 0$
$y = x + \dfrac{1}{x}$ for $x \neq 0$
$y = x^2 + 3$
$y = x$
$y = \dfrac{1}{x^3}$ for $x \neq 0$
This is an Underground Mathematics resource
Student Solutions
Well done to Pablo and Sergio from Kings College of Alicante and Jay from St Stephens Carramar who both found ways to complete the two-way function table.
Here is Pablo's and Sergio's solution.
| $y$-axis is an asymptote | Symmetrical about $x=1$ | Intercepts the y-axis at 3 | passes through origin | |
|---|---|---|---|---|
| $x = 1$ is a root | $y=\ln{x}$ | $y = \vert x-1\vert$ | $y = -3x+3$ | $y=x^2(x-1)$ |
| has exactly two roots | $y=\frac{1}{x^2}-3$ for $x \neq 0 $ | $y = x(x-2)$ | $y=-x^2+2x+3$ | $y=x(x-5)^2$ |
| x-axis is an asymptote | $y = \dfrac{1}{x}$ for $x \neq 0$ | $y = \dfrac{1}{(x-1)^2}$ for $x \neq 1$ | $y = \dfrac{3}{x+1}$ for $x \neq -1$ | $y=\frac{x}{(x-1)^2}$ for $x \neq 1$ |
| $y \to \infty$ as $x \to \infty$ | $y=\frac{x^2+1}{x}$ for $x \neq 0$ | $y=(x-1)^2$ | $y = 2+(x-1)^4$ | $y=x$ |
Here is Jay's solution.
| $y$-axis is an asymptote | Line of symmetry is $x = 1$ | Y intercept = (0,3) | passes through origin | |
|---|---|---|---|---|
| $x = 1$ is a root | $y = \frac{1}{x}-1$ | $y = \vert x-1\vert$ | $y = -3x+3$ | $y=(2x-1)^2-1$ |
| has exactly two roots | $y = (\frac{1}{x}-2)(\frac{1}{x}+2)$ | $y = x(x-2)$ | $y=(x-3)^2-6$ | $y=(2x-1)^2-1$ |
| x-axis is an asymptote | $y = \dfrac{1}{x}$ for $x \neq 0$ | $y = \dfrac{1}{(x-1)^2}$ for $x \neq 1$ | $y = \dfrac{3}{x+1}$ for $x \neq -1$ | $y=(\frac{1}{x+1}-1)(\frac{1}{x-1})$ |
| $y \to \infty$ as $x \to \infty$ | $y = \frac{2^x}{x}$ | $y = (x-1)^2$ | $y = 2+(x-1)^4$ | $y=x^2$ |