Equivalent representations

For polynomials and rational expressions, equivalence means: the same outputs at every shared input. But different algebraic forms make different features visible at sight:

1. Factored form shows the zeros and their multiplicities: $(x-2)(x+5)$ hands you 2 and $-5$ with no work.
2. Standard (expanded) form shows the leading term, hence end behavior, and the constant term, hence the y-intercept.
3. Divided form (quotient plus remainder over divisor) shows a rational function's end-behavior model: the polynomial it imitates when inputs are huge.

Choosing the form is choosing what you want to see. Exam questions reward reading the feature from the right form instead of grinding every problem back to standard form.

Any rational function can be rewritten by polynomial long division: $$\dfrac{f(x)}{g(x)} = q(x) + \dfrac{r(x)}{g(x)},$$ where the remainder $r$ has smaller degree than $g$. As $|x|$ grows, the leftover fraction $r/g$ shrinks toward 0, so the function behaves like the quotient $q$: the end-behavior model.

Example: $\dfrac{x^2 + 2x + 3}{x + 1} = (x + 1) + \dfrac{2}{x+1}$. For large $|x|$ the $\frac{2}{x+1}$ term fades and the graph hugs the slant asymptote $y = x + 1$. The degree bookkeeping falls out automatically: numerator degree exactly one more than denominator degree gives a linear quotient (slant asymptote); equal degrees give a constant quotient (horizontal asymptote at the ratio of leading coefficients); numerator degree smaller gives quotient 0 (horizontal asymptote $y = 0$).

The memorized degree rules fail people because they get applied as incantations. Derived from division, they cannot be misremembered:

1. $\dfrac{3x^2 + 1}{x^2 - 4}$: equal degrees, quotient constant $= 3$, horizontal asymptote $y = 3$ (leading coefficients, not constant terms).
2. $\dfrac{2x + 1}{x^2 + 3}$: top degree smaller, the whole fraction fades, horizontal asymptote $y = 0$: no leading-coefficient division happens.
3. $\dfrac{x^3}{x^2 + 1} = x - \dfrac{x}{x^2+1}$: top degree one larger, slant asymptote $y = x$.

And one folklore rule to bury: a graph can cross its horizontal or slant asymptote in the middle of the picture. The asymptote is a claim about the far ends only; $\frac{x}{x^2+1}$ crosses $y = 0$ at the origin and still settles onto it both directions out.

Rewriting a fraction only preserves the function if you cancel factors: whole multiplied pieces of numerator and denominator. $\dfrac{x^2 + 3x}{x} = \dfrac{x(x+3)}{x} = x + 3$ for $x \ne 0$: legal, because $x$ multiplies the entire top. Slashing the $x^2$ against the $x$ in $\dfrac{x^2 + 5}{x + 5}$ is not algebra; terms joined by $+$ are not cancellable pieces.

Even legal cancellation owes a domain note. $\dfrac{x^2 - 1}{x - 1}$ and $x + 1$ agree at every input but one: the fraction has no value at $x = 1$ (a hole at height 2), the line does. Equivalent representations must match at every shared domain point, so the honest reduction is $x + 1$, $x \ne 1$. Cheap self-test for any claimed equivalence: plug one number into both forms and see if they agree.