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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.
Here is Jay's solution.
Two-way Functions
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 $x \to \infty$ | $y = \frac{2^x}{x}$ | $y = (x-1)^2$ | $y = 2+(x-1)^4$ | $y=x^2$ |
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