Rational functions and graphs
| English | Chinese | Pinyin |
|---|---|---|
| polynomial | 多项式 | duō xiàng shì |
| vertical asymptote | 垂直渐近线 | chuí zhí jiàn jìn xiàn |
| oblique asymptote | 斜渐近线 | xié jiàn jìn xiàn |
| denominator | 分母 | fēn mǔ |
| rational function | 有理函数 | yǒu lǐ hán shù |
| remainder | 余数 | yú shù |
| quotient | 商 | shāng |
The graph that breaks the rules
- Most functions you've met are polynomials 多项式 — smooth, continuous, defined everywhere. But divide one polynomial by another and you get something wilder: vertical asymptotes 垂直渐近线, oblique asymptotes 斜渐近线, and behaviour that changes dramatically near the zeros of the denominator 分母.
- Rational functions 有理函数 are the gateway to understanding more complex curves.
What is a rational function?
- A rational function is a fraction of two polynomials: $f(x) = \dfrac{P(x)}{Q(x)}$.
- The denominator's zeros give vertical asymptotes (where the function shoots to $\pm\infty$).
Worked example. $f(x) = \dfrac{x^2 - 1}{x - 2}$. Vertical asymptote at $x = 2$. Dividing: $f(x) = x + 2 + \dfrac{3}{x-2}$. The oblique asymptote is $y = x + 2$.
Rational functions
y = a/(x − b) + c
A rational function has asymptotes the curve approaches but never touches.
A rational function is a fraction of two polynomials.
That is the definition — one polynomial over another.
Oblique (slant) asymptotes
- When the top has higher degree than the bottom, the graph has an oblique (slant) asymptote — find it by polynomial long division.
- The remainder 余数 term vanishes as $x \to \infty$, leaving the quotient 商 as the asymptote.
The curve hugs the vertical asymptote x = 2 and the oblique asymptote y = x + 2 far from the origin.
Don't forget the remainder. When dividing, the function equals the quotient plus the remainder over the divisor. The asymptote is just the quotient — the remainder tells you which side of the asymptote the curve approaches from.
A rational function has an oblique (slant) asymptote when the degree of the top is:
When the numerator's degree exceeds the denominator's, dividing out leaves a slant line the curve approaches.
You find an oblique asymptote by:
Polynomial division gives the straight-line part the curve approaches far from the origin.
For f(x) = (x²−1)/(x−2), divide to get f(x) = x + 2 + 3/(x−2). The oblique asymptote is y = x + ?
The quotient is x + 2, so the oblique asymptote is y = x + 2.
Related graphs
- You should relate $y = f(x)$ to:
- $y^2 = f(x)$ — only exists where $f(x) \geq 0$, symmetric about the $x$-axis.
- $y = \dfrac{1}{f(x)}$ — zeros become asymptotes, asymptotes become zeros.
- $y = |f(x)|$ — reflects negative parts above the $x$-axis.
- $y = f(|x|)$ — reflects the right half to the left.
For y = 1/f(x), the zeros of f(x) become vertical asymptotes.
Where f(x) = 0, 1/f(x) is undefined — giving a vertical asymptote.
The graph of y = |f(x)| is obtained from y = f(x) by:
The modulus reflects any part of the graph below the x-axis to above it.
Worked example — sketching $y = \dfrac{1}{f(x)}$
- If $f(x) = x^2 - 4$ (roots at $\pm 2$, minimum at $(0, -4)$):
- $\dfrac{1}{f(x)}$ has vertical asymptotes at $x = \pm 2$.
- The minimum of $f$ at $-4$ becomes a local maximum of $\dfrac{1}{f}$ at $-\dfrac{1}{4}$.
- Find the set of values a rational function takes using the discriminant, and its turning points.
You've got it
- a rational function = polynomial ÷ polynomial
- top degree > bottom degree → an oblique asymptote (found by division)
- learn the related graphs: $y^2 = f$, $1/f$, $|f|$, $f(|x|)$