Course Review

Study Sheet

Course Review

Functions, polynomials, exponentials, all of trigonometry, polar & complex numbers, matrices, conics, sequences, and an introduction to limits

Functions and Their Graphs

Concept
Definition, domain & range

A function ff assigns each input xx (the domain) exactly one output y=f(x)y=f(x); the outputs form the range. Passes the vertical line test.

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  • Polynomial: domain (,)(-\infty,\infty).
  • Radical g(x)\sqrt{g(x)}: require g(x)0g(x)\ge 0.
  • Rational p(x)q(x)\dfrac{p(x)}{q(x)}: require q(x)0q(x)\ne 0.

Symmetry: even if f(x)=f(x)f(-x)=f(x) (yy-axis); odd if f(x)=f(x)f(-x)=-f(x) (origin).

Concept
Difference quotient, rate of change, composition, inverses

Difference quotient: f(x+h)f(x)h, h0\dfrac{f(x+h)-f(x)}{h},\ h\ne 0.   Average rate of change on [a,b][a,b]: f(b)f(a)ba\dfrac{f(b)-f(a)}{b-a} (secant slope).

Combinations: (f±g)(x)=f(x)±g(x)(f\pm g)(x)=f(x)\pm g(x), (fg)(x)=f(x)g(x)(fg)(x)=f(x)g(x), (fg)(x)=f(x)g(x)\left(\tfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}.

Composition: (fg)(x)=f(g(x))(f\circ g)(x)=f\big(g(x)\big) (not commutative). Domain: those xx in dom gg with g(x)g(x) in dom ff.

Inverse: exists iff ff is one-to-one (horizontal line test). Swap x,yx,y and solve; f(f1(x))=xf\big(f^{-1}(x)\big)=x and f1(f(x))=xf^{-1}\big(f(x)\big)=x; graphs reflect over y=xy=x.

Concept
Parent functions & transformations of y=af(bxc)+dy=a\,f(bx-c)+d

Parents: c, x, x, x, x2, x3, 1x, xc,\ x,\ |x|,\ \sqrt{x},\ x^2,\ x^3,\ \dfrac1x,\ \lfloor x\rfloor.

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  • +d+d up / d-d down; f(xc)f(x-c) right cc, f(x+c)f(x+c) left (horizontal moves oppose the sign).
  • a<0a<0: reflect over xx-axis; a>1|a|>1 vertical stretch, 0<a<10<|a|<1 shrink.
  • b<0b<0: reflect over yy-axis; b>1|b|>1 horizontal shrink, 0<b<10<|b|<1 stretch (factor: f(b(xcb))f\big(b(x-\tfrac cb)\big)).
Example
Domain, composition, inverse

f(x)=x+2x3f(x)=\dfrac{\sqrt{x+2}}{x-3}: need x2x\ge -2 and x3x\ne 3, so domain [2,3)(3,)[-2,3)\cup(3,\infty).

f(x)=2x+6f(x)=2x+6\Rightarrow swap x=2y+6f1(x)=x62x=2y+6\Rightarrow f^{-1}(x)=\dfrac{x-6}{2}. Check: 2x62+6=x. 2\cdot\tfrac{x-6}{2}+6=x.\ \checkmark

Tip

Tip: Horizontal transformations and shifts always act opposite to their sign; vertical ones act as written. Restrict the domain (e.g. x0x\ge0 for x2x^2) to force a one-to-one function before inverting.

Polynomial and Rational Functions

Concept
End behavior & zeros

For leading term anxna_nx^n: if nn even, both ends same way (an>0a_n>0 up/up, an<0a_n<0 down/down); if nn odd, ends opposite (an>0a_n>0 down/up). A zero of multiplicity mm: graph crosses if mm odd, touches (turns) if mm even. A degree-nn polynomial has at most n1n-1 turning points.

Concept
Division theorems & rational zeros

Division algorithm: f(x)=d(x)q(x)+r(x)f(x)=d(x)q(x)+r(x).

Remainder Theorem: dividing ff by (xk)(x-k) leaves remainder f(k)f(k).

Factor Theorem: (xk)(x-k) is a factor     f(k)=0\iff f(k)=0.

Rational Zero Test: any rational zero of a polynomial with integer coefficients has the form pq\dfrac{p}{q}, pp\mid (constant term), qq\mid (leading coefficient).

Fundamental Thm of Algebra: degree nn (n1n\ge1) has exactly nn complex zeros (with multiplicity). Complex zeros of a real polynomial come in conjugate pairs a±bia\pm bi.

Reminder — Remainder and Factor theorems:P(x)=(xc)Q(x)+P(c)(xc)P(x)    P(c)=0P(x)=(x-c)Q(x)+P(c) \quad\Longrightarrow\quad (x-c)\mid P(x) \iff P(c)=0
Concept
Rational functions f(x)=N(x)D(x)f(x)=\dfrac{N(x)}{D(x)}
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  • Vertical asymptote at zeros of DD not cancelled by NN; a cancelled factor gives a hole.
  • Horizontal asymptote: deg N<N< deg Dy=0D\Rightarrow y=0; equal degree y=lead Nlead D\Rightarrow y=\dfrac{\text{lead }N}{\text{lead }D}; deg N>N> deg DD\Rightarrow none.
  • Slant asymptote when deg N=N= deg D+1D+1: the quotient from long division.
Example
Synthetic division by (x2)(x-2)

Divide x34x2+x+6x^3-4x^2+x+6 by (x2)(x-2):

214162461230\begin{array}{r|rrrr} 2 & 1 & -4 & 1 & 6\\ & & 2 & -4 & -6\\ \hline & 1 & -2 & -3 & 0 \end{array}

Remainder 00, so f(2)=0f(2)=0 and x34x2+x+6=(x2)(x22x3)=(x2)(x3)(x+1)x^3-4x^2+x+6=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1).

Tip

Tip: A remainder of 00 in synthetic division means both “(xk)(x-k) is a factor” and “f(k)=0f(k)=0.” Use it to peel off one zero at a time until the quotient is quadratic, then factor or use the quadratic formula.

Reminder — The quadratic formula:x=b±b24ac2a(a0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)

Exponential and Logarithmic Functions

Concept
Exponential functions and ee

f(x)=abxf(x)=a\,b^{x} (b>0, b1b>0,\ b\ne1): horizontal asymptote y=0y=0, domain (,)(-\infty,\infty), range (0,)(0,\infty); grows if b>1b>1, decays if 0<b<10<b<1. The natural base e2.71828e\approx2.71828.

Compound interest: A=P ⁣(1+rn)ntA=P\!\left(1+\dfrac{r}{n}\right)^{nt}; continuous: A=PertA=Pe^{rt}.

Concept
Logarithms & properties

y=logbx    by=xy=\log_b x \iff b^{y}=x (so log\log is the inverse of bxb^{x}); lnx=logex\ln x=\log_e x. Domain (0,)(0,\infty).

logb(MN)=logbM+logbNlogb ⁣MN=logbMlogbNlogb(Mp)=plogbMlogbb=1,  logb1=0\begin{aligned} \log_b(MN)&=\log_b M+\log_b N & \log_b\!\frac{M}{N}&=\log_b M-\log_b N\\ \log_b(M^{p})&=p\log_b M & \log_b b&=1,\ \ \log_b 1=0 \end{aligned}

Inverse: blogbx=xb^{\log_b x}=x, logbbx=x\log_b b^{x}=x.   Change of base: logbx=lnxlnb=logxlogb\log_b x=\dfrac{\ln x}{\ln b}=\dfrac{\log x}{\log b}.

Example
Solving exponential & logarithmic equations

3e2x=24e2x=82x=ln8x=12ln8=ln223e^{2x}=24\Rightarrow e^{2x}=8\Rightarrow 2x=\ln 8\Rightarrow x=\tfrac12\ln 8=\ln 2\sqrt2.

log2(x)+log2(x2)=3log2 ⁣(x(x2))=3x22x=8(x4)(x+2)=0\log_2(x)+\log_2(x-2)=3\Rightarrow \log_2\!\big(x(x-2)\big)=3\Rightarrow x^2-2x=8\Rightarrow (x-4)(x+2)=0. Reject x=2x=-2 (domain), so x=4x=4.

Tip

Tip: Condense to a single log before exponentiating, or take a log of both sides to bring an exponent down. Always check solutions against the domain x>0x>0 of every logarithm; discard extraneous roots.

Trigonometric Functions

Concept
Angles, radians & arc length

180=π180^\circ=\pi rad. Arc length: s=rθs=r\theta; area of sector: A=12r2θA=\tfrac12 r^2\theta; angular/linear speed: v=rωv=r\omega (θ\theta in radians). Coterminal: add multiples of 360360^\circ or 2π2\pi.

Concept
The six functions & unit circle

On the unit circle a terminal point is (cosθ,sinθ)(\cos\theta,\sin\theta); tanθ=sinθcosθ\tan\theta=\dfrac{\sin\theta}{\cos\theta}, and csc,sec,cot\csc,\sec,\cot are the reciprocals of sin,cos,tan\sin,\cos,\tan. Right triangle: sin=opphyp\sin=\dfrac{\text{opp}}{\text{hyp}}, cos=adjhyp\cos=\dfrac{\text{adj}}{\text{hyp}}, tan=oppadj\tan=\dfrac{\text{opp}}{\text{adj}}.

First-quadrant exact values:

θ0π6π4π3π2sin01222321cos13222120tan03313und.\begin{array}{c|ccccc} \theta & 0 & \tfrac{\pi}{6} & \tfrac{\pi}{4} & \tfrac{\pi}{3} & \tfrac{\pi}{2}\\ \hline \sin & 0 & \tfrac12 & \tfrac{\sqrt2}{2} & \tfrac{\sqrt3}{2} & 1\\[2pt] \cos & 1 & \tfrac{\sqrt3}{2} & \tfrac{\sqrt2}{2} & \tfrac12 & 0\\[2pt] \tan & 0 & \tfrac{\sqrt3}{3} & 1 & \sqrt3 & \text{und.} \end{array}

Signs by quadrant: ASTC — All, Sine, Tangent, Cosine positive in QI--QIV.

Concept
Graphs & inverse trig

For y=asin(bxc)+dy=a\sin(bx-c)+d (same for cos\cos): amplitude a|a|, period 2πb\dfrac{2\pi}{b}, phase shift cb\dfrac{c}{b}, midline y=dy=d. For tan\tan the period is πb\dfrac{\pi}{b}.

Inverse ranges: arcsinx[π2,π2]\arcsin x\in[-\tfrac\pi2,\tfrac\pi2], arccosx[0,π]\arccos x\in[0,\pi], arctanx(π2,π2)\arctan x\in(-\tfrac\pi2,\tfrac\pi2).

Example
Reading a sine graph

y=3sin ⁣(2xπ2)+1y=3\sin\!\big(2x-\tfrac\pi2\big)+1: amplitude 33, period 2π2=π\dfrac{2\pi}{2}=\pi, phase shift π/22=π4\dfrac{\pi/2}{2}=\dfrac\pi4 right, midline y=1y=1 (so yy ranges from 2-2 to 44).

Tip

Tip: Use a reference angle (acute angle to the xx-axis) plus ASTC to get exact values in any quadrant. Memorize QI values; every other value is one of them with a sign attached.

Analytic Trigonometry

Concept
Fundamental identities

Reciprocal/Quotient: csc=1sin\csc=\tfrac1{\sin}, sec=1cos\sec=\tfrac1{\cos}, cot=1tan=cossin\cot=\tfrac1{\tan}=\tfrac{\cos}{\sin}, tan=sincos\tan=\tfrac{\sin}{\cos}.

Pythagorean: sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1, 1+tan2θ=sec2θ1+\tan^2\theta=\sec^2\theta, 1+cot2θ=csc2θ1+\cot^2\theta=\csc^2\theta.

Even/Odd: sin(θ)=sinθ\sin(-\theta)=-\sin\theta, cos(θ)=cosθ\cos(-\theta)=\cos\theta, tan(θ)=tanθ\tan(-\theta)=-\tan\theta.

Cofunction: sin ⁣(π2θ)=cosθ\sin\!\big(\tfrac\pi2-\theta\big)=\cos\theta, tan ⁣(π2θ)=cotθ\tan\!\big(\tfrac\pi2-\theta\big)=\cot\theta.

Concept
Sum/difference, double, half, power-reducing, product-to-sum
sin(A±B)=sinAcosB±cosAsinBcos(A±B)=cosAcosBsinAsinBtan(A±B)=tanA±tanB1tanAtanBsin2θ=2sinθcosθcos2θ=cos2θsin2θ=2cos2θ1=12sin2θtan2θ=2tanθ1tan2θ\begin{aligned} \sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B\\ \cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B\\ \tan(A\pm B)&=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}\\ \sin2\theta&=2\sin\theta\cos\theta\\ \cos2\theta&=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta\\ \tan2\theta&=\frac{2\tan\theta}{1-\tan^2\theta} \end{aligned}

Power-reducing: sin2θ=1cos2θ2\sin^2\theta=\dfrac{1-\cos2\theta}{2}, cos2θ=1+cos2θ2\cos^2\theta=\dfrac{1+\cos2\theta}{2}.

Half-angle: sinθ2=±1cosθ2\sin\dfrac\theta2=\pm\sqrt{\dfrac{1-\cos\theta}{2}}, cosθ2=±1+cosθ2\cos\dfrac\theta2=\pm\sqrt{\dfrac{1+\cos\theta}{2}}, tanθ2=1cosθsinθ=sinθ1+cosθ\tan\dfrac\theta2=\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{\sin\theta}{1+\cos\theta}.

Product-to-sum: sinAsinB=12[cos(AB)cos(A+B)]\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)], cosAcosB=12[cos(AB)+cos(A+B)]\cos A\cos B=\tfrac12[\cos(A-B)+\cos(A+B)], sinAcosB=12[sin(A+B)+sin(AB)]\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)].

Example
Exact value & solving

cos15=cos(4530)=2232+2212=6+24\cos15^\circ=\cos(45^\circ-30^\circ)=\tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2}+\tfrac{\sqrt2}{2}\cdot\tfrac12=\dfrac{\sqrt6+\sqrt2}{4}.

2cos2xcosx1=0(2cosx+1)(cosx1)=0cosx=122\cos^2x-\cos x-1=0\Rightarrow(2\cos x+1)(\cos x-1)=0\Rightarrow\cos x=-\tfrac12 or 11; on [0,2π)[0,2\pi): x=0,2π3,4π3x=0,\tfrac{2\pi}{3},\tfrac{4\pi}{3}.

Tip

Tip: To verify an identity, work one side only — convert to sin/cos\sin/\cos, get a common denominator, and apply a Pythagorean identity. To solve, factor to one function and remember to add +2πn+2\pi n (sin,cos\sin,\cos) or +πn+\pi n (tan\tan) for all solutions.

Reminder — The Pythagorean identity:sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1

Triangle Trigonometry & Vectors

Concept
Laws of Sines and Cosines, area

Law of Sines: asinA=bsinB=csinC\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} (cases AAS, ASA, SSA).

Ambiguous case (SSA): given a,b,Aa,b,A with AA acute and h=bsinAh=b\sin A: no triangle if a<ha<h; one if a=ha=h or aba\ge b; two if h<a<bh<a<b.

Law of Cosines: c2=a2+b22abcosCc^2=a^2+b^2-2ab\cos C (cases SAS, SSS).

Area: 12absinC\tfrac12 ab\sin C; Heron: s(sa)(sb)(sc)\sqrt{s(s-a)(s-b)(s-c)} with s=a+b+c2s=\dfrac{a+b+c}{2}.

Reminder — Law of Sines:asinA=bsinB=csinC\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Reminder — Law of Cosines:c2=a2+b22abcosCc^{2}=a^{2}+b^{2}-2ab\cos C
Concept
Vectors, dot product & projection

v=v1,v2\mathbf{v}=\langle v_1,v_2\rangle, magnitude v=v12+v22\|\mathbf{v}\|=\sqrt{v_1^2+v_2^2}, direction angle θ\theta with v=vcosθ,sinθ\mathbf{v}=\|\mathbf{v}\|\langle\cos\theta,\sin\theta\rangle. Unit vector u=vv\mathbf{u}=\dfrac{\mathbf{v}}{\|\mathbf{v}\|}; standard form v1i+v2jv_1\mathbf{i}+v_2\mathbf{j}.

Dot product: uv=u1v1+u2v2=uvcosθ\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=\|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta. Angle: cosθ=uvuv\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}. Orthogonal     uv=0\iff\mathbf{u}\cdot\mathbf{v}=0.

Projection: projvu=uvv2v\operatorname{proj}_{\mathbf{v}}\mathbf{u}=\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}.

Example
Law of Cosines (SAS)

a=5, b=8, C=60a=5,\ b=8,\ C=60^\circ: c2=25+642(5)(8)cos60=898012=49c^2=25+64-2(5)(8)\cos60^\circ=89-80\cdot\tfrac12=49, so c=7c=7. Area =12(5)(8)sin60=103=\tfrac12(5)(8)\sin60^\circ=10\sqrt3.

Tip

Tip: Choose the Law of Cosines when you know all three sides (SSS) or two sides and the included angle (SAS); otherwise use the Law of Sines. In the SSA case always test whether a second triangle (supplementary angle) also fits.

Polar Coordinates & Complex Numbers

Concept
Polar & rectangular conversion

Point (r,θ)(r,\theta). Polar \to rectangular: x=rcosθ, y=rsinθx=r\cos\theta,\ y=r\sin\theta. Rectangular \to polar: r=x2+y2, tanθ=yxr=\sqrt{x^2+y^2},\ \tan\theta=\dfrac yx (choose θ\theta by the quadrant of (x,y)(x,y)).

Common curves: circle r=ar=a; line θ=α\theta=\alpha; rose r=acos(nθ)r=a\cos(n\theta) (nn petals if odd, 2n2n if even); limaçon r=a±bcosθr=a\pm b\cos\theta; cardioid when a=ba=b.

Concept
Complex numbers & trig form

z=a+biz=a+bi, z=r=a2+b2|z|=r=\sqrt{a^2+b^2}. Trig (polar) form: z=r(cosθ+isinθ)z=r(\cos\theta+i\sin\theta) with tanθ=ba\tan\theta=\dfrac ba.

Product/Quotient: z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]z_1z_2=r_1r_2\big[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big], z1z2=r1r2[cos(θ1θ2)+isin(θ1θ2)]\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\big[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\big].

DeMoivre: zn=rn(cosnθ+isinnθ)z^n=r^n\big(\cos n\theta+i\sin n\theta\big).

nnth roots: rn[cosθ+2πkn+isinθ+2πkn]\sqrt[n]{r}\left[\cos\dfrac{\theta+2\pi k}{n}+i\sin\dfrac{\theta+2\pi k}{n}\right], k=0,1,,n1k=0,1,\dots,n-1 (equally spaced by 2πn\dfrac{2\pi}{n}).

Example
DeMoivre's Theorem

(1+i)8\big(1+i\big)^{8}: here r=2, θ=45r=\sqrt2,\ \theta=45^\circ, so (1+i)8=(2)8(cos360+isin360)=16(1+0i)=16(1+i)^8=(\sqrt2)^8\big(\cos360^\circ+i\sin360^\circ\big)=16(1+0i)=16.

Tip

Tip: Convert to trig form before raising to a power or taking a root — DeMoivre turns messy multiplication into multiplying moduli and adding angles. The nn distinct nnth roots all share modulus rn\sqrt[n]{r} and sit at equal angular spacing.

Systems, Matrices & Determinants

Concept
Solving systems & Gaussian elimination

Methods: substitution, elimination, or row-reduce the augmented matrix to row-echelon form (leading 1's, zeros below) then back-substitute. Row operations: swap rows, multiply a row by a nonzero constant, add a multiple of one row to another. A system is inconsistent (no solution) if a row gives 0=0= nonzero; dependent (infinitely many) if a variable is free.

Concept
Matrix operations, inverse, determinants, Cramer

Add/subtract entrywise (same size). Product ABAB: rows of AA dotted with columns of BB; defined when (cols of AA) == (rows of BB); not commutative. Identity II: AI=IA=AAI=IA=A.

2×22\times2 inverse: for A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, detA=adbc\det A=ad-bc and A1=1adbc[dbca]A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix} (exists iff detA0\det A\ne0).

3×33\times3 determinant by cofactor expansion. Cramer's Rule: xi=detAidetAx_i=\dfrac{\det A_i}{\det A}, where AiA_i replaces column ii with the constants (requires detA0\det A\ne0).

Concept
Partial fraction decomposition

For a proper rational N(x)D(x)\dfrac{N(x)}{D(x)} (divide first if improper), split by the factors of DD:

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  • Linear (ax+b)(ax+b): term Aax+b\dfrac{A}{ax+b}; repeated (ax+b)k(ax+b)^k: one term per power up to kk.
  • Irreducible quadratic (ax2+bx+c)(ax^2+bx+c): term Bx+Cax2+bx+c\dfrac{Bx+C}{ax^2+bx+c}.

Clear denominators and match coefficients (or substitute convenient xx-values) to solve.

Example
2×22\times2 inverse and Cramer

A=[2153]A=\begin{bmatrix}2&1\\5&3\end{bmatrix}, detA=2(3)1(5)=1\det A=2(3)-1(5)=1, so A1=[3152]A^{-1}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}.

For {2x+y=55x+3y=13\begin{cases}2x+y=5\\5x+3y=13\end{cases}: x=det[51133]1=15131=2x=\dfrac{\det\begin{bmatrix}5&1\\13&3\end{bmatrix}}{1}=\dfrac{15-13}{1}=2, y=det[25513]1=2625=1y=\dfrac{\det\begin{bmatrix}2&5\\5&13\end{bmatrix}}{1}=26-25=1.

Tip

Tip: An inverse (and a unique solution via Cramer) exists only when detA0\det A\ne0. If detA=0\det A=0 the system is either inconsistent or dependent — fall back on row reduction to tell which.

Conic Sections & Parametric Equations

Concept
Standard forms (center/vertex (h,k)(h,k))

Circle: (xh)2+(yk)2=r2(x-h)^2+(y-k)^2=r^2.

Parabola: (xh)2=4p(yk)(x-h)^2=4p(y-k) opens up/down; (yk)2=4p(xh)(y-k)^2=4p(x-h) opens left/right. Focus is pp from vertex; directrix is the opposite side; 4p|4p| is the latus rectum.

Ellipse: (xh)2a2+(yk)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1, a>ba>b: major axis length 2a2a, minor 2b2b, foci cc from center with c2=a2b2c^2=a^2-b^2.

Hyperbola: (xh)2a2(yk)2b2=1\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1: vertices aa from center, foci with c2=a2+b2c^2=a^2+b^2, asymptotes yk=±ba(xh)y-k=\pm\dfrac ba(x-h).

Concept
Eccentricity & parametric equations

Eccentricity e=cae=\dfrac ca: circle e=0e=0; ellipse 0<e<10<e<1; parabola e=1e=1; hyperbola e>1e>1.

Parametric: x=f(t), y=g(t)x=f(t),\ y=g(t) traces a curve with orientation as tt increases. Eliminate the parameter by solving for tt (or using cos2t+sin2t=1\cos^2t+\sin^2t=1) to get a relation in x,yx,y.

Example
Complete the square; eliminate the parameter

x2+4y24x=0(x2)2+4y2=4(x2)24+y2=1x^2+4y^2-4x=0\Rightarrow(x-2)^2+4y^2=4\Rightarrow\dfrac{(x-2)^2}{4}+y^2=1: ellipse, center (2,0)(2,0), a=2, b=1a=2,\ b=1.

x=2cost, y=3sint(x2)2+(y3)2=cos2t+sin2t=1x=2\cos t,\ y=3\sin t\Rightarrow\left(\tfrac x2\right)^2+\left(\tfrac y3\right)^2=\cos^2t+\sin^2t=1: ellipse x24+y29=1\dfrac{x^2}{4}+\dfrac{y^2}{9}=1.

Tip

Tip: Read the conic from the signs: two squared terms added with equal denominators == circle, unequal == ellipse; subtracted == hyperbola; only one squared term == parabola. Complete the square to reach standard form.

Sequences, Series & the Binomial Theorem

Concept
Arithmetic & geometric

Arithmetic (common difference dd): an=a1+(n1)da_n=a_1+(n-1)d; sum Sn=n2(a1+an)=n2[2a1+(n1)d]S_n=\dfrac n2(a_1+a_n)=\dfrac n2\big[2a_1+(n-1)d\big].

Geometric (common ratio rr): an=a1rn1a_n=a_1r^{\,n-1}; sum Sn=a11rn1rS_n=a_1\dfrac{1-r^{n}}{1-r}.

Infinite geometric: converges iff r<1|r|<1, then S=a11rS=\dfrac{a_1}{1-r}.

Concept
Sigma notation, induction, counting

Sigma: i=1nai=a1+a2++an\displaystyle\sum_{i=1}^{n}a_i=a_1+a_2+\cdots+a_n. Useful: i=1ni=n(n+1)2\displaystyle\sum_{i=1}^{n}i=\dfrac{n(n+1)}{2}.

Mathematical induction: prove a base case (n=1n=1), then assume P(k)P(k) and prove P(k+1)P(k+1); concludes P(n)P(n) for all nn.

Counting: permutations nPr=n!(nr)!_nP_r=\dfrac{n!}{(n-r)!} (order matters); combinations nCr=(nr)=n!r!(nr)!_nC_r=\dbinom{n}{r}=\dfrac{n!}{r!(n-r)!} (order does not).

Concept
Binomial Theorem
(a+b)n=k=0n(nk)ankbk.(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{k}.

The (k+1)(k+1)th term is (nk)ankbk\dbinom{n}{k}a^{\,n-k}b^{k}. Coefficients form Pascal's triangle; (nk)=(nnk)\dbinom{n}{k}=\dbinom{n}{n-k}.

Reminder — The binomial theorem:(x+y)n=k=0n(nk)xnkyk(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{\,n-k}y^{k}
Example
Series values & expansion

Infinite geometric 3+1+13+3+1+\tfrac13+\cdots: r=13r=\tfrac13, so S=3113=323=92S=\dfrac{3}{1-\frac13}=\dfrac{3}{\frac23}=\dfrac92.

(x+2)4=x4+4x3(2)+6x2(4)+4x(8)+16=x4+8x3+24x2+32x+16(x+2)^4=x^4+4x^3(2)+6x^2(4)+4x(8)+16=x^4+8x^3+24x^2+32x+16.

Tip

Tip: Check for a common difference (arithmetic) versus a common ratio (geometric) before choosing a formula. An infinite geometric series has a finite sum only when r<1|r|<1.

Introduction to Limits

Concept
Meaning & limit laws

limxcf(x)=L\displaystyle\lim_{x\to c}f(x)=L means f(x)f(x) approaches LL as xcx\to c. It exists iff the one-sided limits agree: limxcf=limxc+f\displaystyle\lim_{x\to c^-}f=\lim_{x\to c^+}f. Limits ignore the actual value f(c)f(c).

For constants/sums/products/quotients the limit distributes:

lim(f±g)=limf±limg,lim(fg)=limflimg,limfg=limflimg (limg0).\lim(f\pm g)=\lim f\pm\lim g,\quad \lim(fg)=\lim f\cdot\lim g,\quad \lim\frac fg=\frac{\lim f}{\lim g}\ (\lim g\ne0).

Polynomials/rationals: substitute cc when the denominator is nonzero.

Concept
Indeterminate forms, infinite limits, limits at infinity

00\tfrac00 form: factor and cancel, rationalize, or simplify, then substitute.

Infinite limits: if f±f\to\pm\infty near cc there is a vertical asymptote x=cx=c.

Limits at infinity (end behavior of rationals): compare degrees like horizontal asymptotes — lower top 0\to0, equal \to ratio of leading coefficients, higher top ±\to\pm\infty.

Concept
Continuity & the derivative

ff is continuous at cc if all hold: (1) f(c)f(c) defined, (2) limxcf(x)\displaystyle\lim_{x\to c}f(x) exists, (3) limxcf(x)=f(c)\displaystyle\lim_{x\to c}f(x)=f(c).

Limit definition of the derivative:

f(x)=limh0f(x+h)f(x)h,f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h},

the instantaneous rate of change / slope of the tangent line (the difference quotient as h0h\to0).

Example
A 00\tfrac00 limit and a derivative

limx2x24x2=limx2(x2)(x+2)x2=limx2(x+2)=4\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\to2}(x+2)=4.

For f(x)=x2f(x)=x^2: f(x)=limh0(x+h)2x2h=limh0(2x+h)=2x\displaystyle f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}(2x+h)=2x.

Tip

Tip: Try direct substitution first. If you get 00\tfrac00, the factor causing it cancels — factor, rationalize, or simplify, then substitute again. A nonzero-over-zero result signals an infinite limit (vertical asymptote), not 00.