The absolute value of \(x\) is defined as:
This always gives a positive result.
Graph of \(y = 3x^2 + 6x - 2\):
Graph of \(y = |3x^2 + 6x - 2|\):
Note how the negative portions have been reflected in the x‑axis.
\(y = |3 \tan x|\)
\(y = |3x + 2|\)
Odd functions have half‑turn symmetry about the origin:
\[
f(-x) = -f(x)
\]
\(y = x^3\)
\(y = x^5 - 3x\)
Show that \(x^5 + 3x^3\) is an odd function.
Even functions are symmetrical about the y‑axis:
\[
f(-x) = f(x)
\]
\(y = x^4 - 1\)
Is \(x^6 + 3x^2\) an even function?
An asymptote to a curve is a straight line which the curve approaches but never reaches.
\(f(x) = \frac{1}{x}\)
The graph \(y = 1/x\) has vertical asymptote \(x = 0\) and horizontal asymptote \(y = 0\).
To the left of \(x = 0\), \(f(x) \to -\infty\) as \(x \to 0\). To the right of \(x = 0\), \(f(x) \to \infty\) as \(x \to 0\).
\(f(x) = \dfrac{x - 3}{x^3 + 1}\)
Vertical asymptote at \(x = -1\). Horizontal asymptote at \(y = 0\).
Left of \(x = -1\): \(f(x) \to \infty\). Right of \(x = -1\): \(f(x) \to -\infty\).
\(f(x) = \dfrac{(x+1)(x-3)}{(x+3)(x-4)}\)
Vertical asymptotes at \(x = -3\) and \(x = 4\). Horizontal asymptote at \(y = 1\).
Vertical asymptotes occur when the denominator is zero.
Horizontal and oblique asymptotes require algebraic division.
The quotient becomes the asymptote.
Find the asymptotes of:
Alternatively:
Asymptotes parallel to the x‑axis can be found by equating the coefficient of the highest power of \(x\) to zero. Those parallel to the y‑axis are found similarly.
To find oblique asymptotes, substitute \(y = mx + c\) and equate coefficients of the two highest powers of \(x\).
To sketch a function with asymptotes, follow these steps:
Using the example above, sketch:
Vertical asymptotes:
Horizontal/oblique asymptotes:
y‑intercept:
x‑intercept:
Stationary points:
Sketch:
Sketch the function:

Vertical asymptotes:
The y‑intercept:
The x‑intercept:
Stationary points:
Find nature of turning points:
Sketch: