Polynomial and Rational Functions

Standard (vertex) form: $f(x)=a(x-h)^2+k$ has vertex $(h,k)$ and axis of symmetry $x=h$. General form: $f(x)=ax^2+bx+c$ has vertex at $x=-\dfrac{b}{2a}$, so the vertex is $\left(-\dfrac{b}{2a},\,f\!\left(-\dfrac{b}{2a}\right)\right)$. If $a>0$ the parabola opens up (vertex is a minimum); if $a<0$ it opens down (vertex is a maximum).

For $f(x)=a_nx^n+\cdots+a_0$, the ends are controlled by the degree $n$ and sign of $a_n$:

Real zeros & multiplicity: if $(x-c)^k$ is a factor, $c$ is a zero of multiplicity $k$. If $k$ is odd the graph crosses at $c$; if $k$ is even it touches (bounces). A degree-$n$ polynomial has at most $n$ real zeros and at most $n-1$ turning points.

For polynomials, $f(x)=d(x)\,q(x)+r(x)$ with $\deg r<\deg d$. Synthetic division is a shortcut when the divisor is $x-k$. Remainder Theorem: dividing $f(x)$ by $x-k$ leaves remainder $r=f(k)$. Factor Theorem: $(x-k)$ is a factor of $f(x)$ if and only if $f(k)=0$.

Rational Zero Test: every rational zero of $f$ has the form $\dfrac{p}{q}$, where $p\mid(\text{constant term})$ and $q\mid(\text{leading coefficient})$. Descartes' Rule of Signs: the number of positive real zeros equals the number of sign changes in $f(x)$, or less by an even number. The number of negative real zeros uses the sign changes in $f(-x)$. Upper/Lower Bound Rule (via synthetic division by $x-c$): if $c>0$ and the bottom row is all $\ge0$, then $c$ is an upper bound for the real zeros. If $c<0$ and the bottom row alternates in sign, then $c$ is a lower bound.

With $i^2=-1$, a complex number is $a+bi$ and its conjugate is $a-bi$; note $(a+bi)(a-bi)=a^2+b^2$. Fundamental Theorem of Algebra: every polynomial of degree $n\ge1$ has exactly $n$ complex zeros (counting multiplicity). Conjugate Pairs Theorem: if $f$ has real coefficients and $a+bi$ is a zero, then $a-bi$ is also a zero.

For $f(x)=\dfrac{N(x)}{D(x)}$ in lowest terms: Domain: all reals except the zeros of $D(x)$. Vertical asymptotes: at zeros of $D(x)$ that remain after cancelling. Holes: at any zero common to $N$ and $D$ (a cancelled factor). Horizontal / slant asymptotes (let $n=\deg N,\ m=\deg D$):

If $x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ (monic) has roots $r_1,\dots,r_n$, then the elementary symmetric sums of the roots are the coefficients (up to sign):

Cubic $x^3+bx^2+cx+d$: $\ r_1+r_2+r_3=-b$, $\ r_1r_2+r_1r_3+r_2r_3=c$, $\ r_1r_2r_3=-d$. For a non-monic $a_nx^n+\cdots+a_0$, divide by $a_n$ first: e.g. $\sum r_i=-\dfrac{a_{n-1}}{a_n}$ and $\prod r_i=(-1)^n\dfrac{a_0}{a_n}$.

Let $p_k=r_1^k+r_2^k+\cdots+r_n^k$ be the $k$-th power sum of the roots and let $e_1,e_2,\dots$ be the elementary symmetric sums ($e_1=\sum r_i,\ e_2=\sum_{i<j}r_ir_j,\ \dots$). Newton's identities link them:

These let you compute $\sum r_i^k$ without ever finding the roots — just read $e_1,e_2,\dots$ off the coefficients via Vieta.

Dividing $f(x)$ by a quadratic $(x-a)(x-b)$ leaves a remainder of degree $<2$, so $r(x)=px+q$ is linear:

Evaluating at the two roots kills the quotient term, giving a $2\times2$ system:

Solve for $p,q$. (The idea generalizes: the remainder mod a degree-$m$ divisor has degree $<m$, and $m$ evaluation points pin it down — this is polynomial interpolation.)

From constraints: a degree-$n$ polynomial has $n+1$ coefficients, so $n+1$ independent conditions (values, zeros, matching derivatives/slopes, a leading coefficient) determine it. Set up one linear equation per condition and solve. Roots in arithmetic progression (AP): write three roots symmetrically as $\alpha-\delta,\ \alpha,\ \alpha+\delta$. Then $\sum r_i=3\alpha$, so Vieta immediately gives the middle root $\alpha=-\tfrac{b}{3}$ (for monic $x^3+bx^2+cx+d$). Roots in geometric progression (GP): write them as $\tfrac{\alpha}{\rho},\ \alpha,\ \alpha\rho$. Then the product $r_1r_2r_3=\alpha^3=-d$, so the middle root is $\alpha=\sqrt[3]{-d}$.

To split a proper rational function $\dfrac{N(x)}{D(x)}$ (with $\deg N<\deg D$; otherwise divide first), fully factor $D$ and assign one term per factor:

Clear denominators, then solve for the constants by substituting convenient $x$-values (the roots) and/or matching coefficients.

Slant asymptote occurs exactly when $\deg N=\deg D+1$: polynomial-divide $N$ by $D$; the quotient (a line $y=mx+b$) is the slant asymptote and the remainder term $\to0$ as $x\to\pm\infty$. Crossing the horizontal asymptote: an asymptote controls only end behavior — a graph may cross its horizontal (or slant) asymptote at finite $x$. To find where, set $f(x)$ equal to the asymptote value and solve; real solutions are the crossing points. (A graph can never cross a vertical asymptote, but horizontal/slant ones are fair game.)