Or search by topic
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 |
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 |
This comes in two parts, with the first being less fiendish than the second. It’s great for practising both quadratics and laws of indices, and you can get a lot from making sure that you find all the solutions. For a real challenge (requiring a bit more knowledge), you could consider finding the complex solutions.
You're invited to decide whether statements about the number of solutions of a quadratic equation are always, sometimes or never true.
This will encourage you to think about whether all quadratics can be factorised and to develop a better understanding of the effect that changing the coefficients has on the factorised form.