ABC Calculus I - Comprehensive Study Notes (10 Topics)

Fractions

  • Concept: Fractions represent parts of a whole. Denominator indicates the size of the piece, numerator how many pieces. Two fractions can only be added if they count the same size pieces.
  • Technique 1 (common denominator): To add a/b and c/d, rewrite so both have a common denominator, then add numerators. A common denominator can be found by multiplying top and bottom of each fraction by a factor that yields the common denominator.
  • Example 1.1 (1/2 + 2/3):
    • Common denominator: 6
    • Convert: 12=36,  23=46\frac{1}{2} = \frac{3}{6}, \; \frac{2}{3} = \frac{4}{6}
    • Sum: 36+46=76\frac{3}{6} + \frac{4}{6} = \frac{7}{6}
  • Example 1.2 (with variables): 2xx+1+1x−3\frac{2x}{x+1} + \frac{1}{x-3}
    • If the denominators have no common factors, LCD = (x+1)(x-3)
    • Multiply to get common denominator and add:
    • 2xx+1⋅x−3x−3+1x−3⋅x+1x+1=2x(x−3)+(x+1)(x+1)(x−3)\frac{2x}{x+1}\cdot\frac{x-3}{x-3} + \frac{1}{x-3}\cdot\frac{x+1}{x+1} = \frac{2x(x-3) + (x+1)}{(x+1)(x-3)}
    • Numerator: 2x2−6x+x+1=2x2−5x+12x^2 - 6x + x + 1 = 2x^2 - 5x + 1
    • Result: 2x2−5x+1(x+1)(x−3)\frac{2x^2 - 5x + 1}{(x+1)(x-3)}
    • Note: When possible, leave the denominator factored. Why? (1) Factored forms are preferable; (2) less room for mistakes which would cost full credit on the ABC (no partial credit).
  • Example 1.3 (3/t^2 − 3/t):
    • Write with LCD t^2: 3t2−3t=3t2−3tt2=3−3tt2=3(1−t)t2\frac{3}{t^2} - \frac{3}{t} = \frac{3}{t^2} - \frac{3t}{t^2} = \frac{3 - 3t}{t^2} = \frac{3(1 - t)}{t^2}
  • Remarks:
    • In some editions, the least common denominator and product LCDs are discussed; if denominators share factors, you may use a common denominator that’s not the least, then reduce at the end.

Factoring

  • Goal: Express a polynomial as a product of irreducible factors.
  • Technique 2 (simple trinomials): For x^2 + bx + c, find r and s such that r + s = b and rs = c. Then x^2 + bx + c = (x + r)(x + s).
  • Example 2.1: Factor x^2 − 9x − 36
    • Find r, s with r + s = −9 and rs = −36. r = 3, s = −12 → x^2 − 9x − 36 = (x + 3)(x − 12).
  • Technique 3 (a ≠ 1): Factor ax^2 + bx + c by first factoring out gcd, then (AC) method:
    • Multiply a and c; find r, s with r + s = b and rs = ac; rewrite ax^2 + bx + c = (ax^2 + rx) + (sx + c); factor by grouping; look for a common binomial factor.
  • Example 2.2: Factor 6t^2 + 7t − 5
    • ac = (6)(−5) = −30; find r, s with rs = −30, r + s = 7. r = 10, s = −3
    • Rewrite: 6t^2 + 7t − 5 = (6t^2 − 3t) + (10t − 5) = 3t(2t − 1) + 5(2t − 1) = (3t + 5)(2t − 1).
  • Example 2.3: Factor 36x^2 − 18x − 4
    • Common factor 2: 2(18x^2 − 9x − 2)
    • ac = 18(−2) = −36; r, s with rs = −36, r + s = −9 → r = −12, s = 3
    • Group: 2[(18x^2 − 12x) + (3x − 2)] = 2[6x(3x − 2) + 1(3x − 2)] = 2(6x + 1)(3x − 2)
  • Tips:
    • Check by expanding to verify. A correct factorization should multiply back to the original expression.

Solving Equations

  • Goal: Solve for a variable in an equation using inverse operations. If the variable appears only to the first power, isolate with inverse operations; if higher powers, treat as a polynomial or other strategies.
  • Technique 4 (quadratic equations): For ax^2 + bx + c = 0, bring all terms to one side, then either factor or use the quadratic formula x = \frac{−b \pm \sqrt{b^2 − 4ac}}{2a}.
  • Important: Quadratic equations typically have two solutions. If b^2 − 4ac = 0, there is a double root; if b^2 − 4ac < 0, no real solutions.
  • Example 3.1: Solve 2t^2 + 6t − 3 = t
    • Move terms: 2t^2 + 5t − 3 = 0
    • Factor: (2t − 1)(t + 3) = 0 → t = 1/2 or t = −3
    • Quadratic formula: x = (−5 ± sqrt(25 + 24)) / 4 = (−5 ± 7)/4 → x = 1/2 or −3
  • Example 3.2: Solve z^2 = 16
    • z^2 − 16 = 0 → (z − 4)(z + 4) = 0 → z = ±4
  • Technique 5 (clear denominators): When an equation has fractions, multiply both sides by a common denominator to clear denominators. Then solve and check for extraneous solutions (denominators cannot be zero in the original equation).
  • Example 3.3: Solve 5 − 8x = 3/4
    • Clear denominators by multiplying by 4: 4(5 − 8x) = 3 → 20 − 32x = 3 → −32x = −17 → x = 17/32
    • Check: original denominators do not vanish for x = 17/32, so it’s valid.
  • Example 3.4: Solve t + 2/(t − 1) = 2t/(t − 1)
    • Multiply both sides by (t − 1): t(t − 1) + 2 = 2t
    • Simplify: t^2 − t + 2 = 2t → t^2 − 3t + 2 = 0 → (t − 1)(t − 2) = 0 → t = 1 or t = 2
    • Check: t = 1 makes a denominator zero in the original equation, so discard; final t = 2.
  • Technique 6 (square roots): If an equation contains a square root, isolate it first, then square both sides. Be aware that squaring may introduce extraneous solutions.
  • Example 3.5: Solve x = \sqrt{4x + 9} − 1
    • Isolate: x + 1 = \sqrt{4x + 9}
    • Square both sides: x^2 + 2x + 1 = 4x + 9 → x^2 − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2
    • Check in original: x = −2 does not satisfy; only x = 4 is valid.

Inequalities

  • Definition: An inequality uses
  • Rule: If A < B and c > 0, then Ac < Bc; if c < 0, then Ac > Bc (sign flip).
  • Technique 7: Solve inequalities by inverse operations, but flip the inequality sign when multiplying/dividing by a negative number.
  • Example 4.1 (number line intuition with negation): If you multiply both sides of a true inequality by −1, the inequality flips; e.g., if 7 < 8, then −7 > −8.
  • Example 4.2: Solve 3 + x/(3 − x) ≥ 1
    • Treat like an equation but consider sign of the denominator (3 − x)
    • Consider two cases: 3 − x > 0 (x < 3) and 3 − x < 0 (x > 3)
    • After analysis, solution is 0 ≤ x < 3 (note: x cannot be 3 because denominator would be zero)
  • Example 4.3: Solve t − 3t + 1 ≥ 0 (i.e., (t − 3)/(t + 1) ≥ 0)
    • Critical points: t = 3 and t = −1 (denominator zero at t = −1)
    • Sign analysis yields t = 3 or t > 3, or t < −1; combining with the fact that t = −1 is excluded gives t ≥ 3 or t < −1
  • Example 4.4: Absolute value inequality |3r + 2| ≥ 4
    • Solve as two inequalities: 3r + 2 ≥ 4 or 3r + 2 ≤ −4
    • Solutions: r ≥ 2/3 or r ≤ −2
  • Example 4.5: |5 − z| < 3
    • Translate: −3 < 5 − z < 3
    • Solve for z: 2 < z < 8
  • Example 4.6: 32 − 2x^2 ≥ 0
    • Rewrite: −2x^2 ≥ −32 → x^2 ≤ 16 → −4 ≤ x ≤ 4 (also note √x^2 = |x|, so |x| ≤ 4)

The Equation of a Line

  • Key concepts:
    • Slope m of line through (x1, y1) and (x2, y2): m=y<em>2−y</em>1x<em>2−x</em>1m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}
    • Point-slope form: y−y<em>1=m(x−x</em>1)y - y<em>1 = m(x - x</em>1)
    • Slope-intercept form: y=mx+by = mx + b with b=y<em>1−mx</em>1b = y<em>1 - mx</em>1 when you plug a known point.
  • Technique 8: To find a line from given data, first identify a point on the line and the slope. Use point-slope form to obtain the equation; convert to slope-intercept form if requested.
  • Example 5.1: Line through (−1, 2) with slope 6
    • Point-slope: y−2=6(x+1)y - 2 = 6(x + 1)
    • Slope-intercept: y=6x+8y = 6x + 8
  • Example 5.2: Line through (2, 2) and (6, −8)
    • Slope: m=−8−26−2=−104=−52m = \frac{-8 - 2}{6 - 2} = \frac{-10}{4} = -\frac{5}{2}
    • Point-slope: y−2=−52(x−2)y - 2 = -\frac{5}{2}(x - 2)
    • Slope-intercept: y=−52x+7y = -\frac{5}{2}x + 7
    • Note: The transcript contains a sign error in the intermediate arithmetic; the correct slope is -5/2 and the correct intercept is 7.

Right Triangles

  • Core relationships:
    • Pythagorean theorem: x2+y2=h2x^2 + y^2 = h^2
    • Similar triangles give trig ratios for a given angle θ in a right triangle:
    • sin⁡θ=oppositehypotenuse=yh\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{h}
    • cos⁡θ=adjacenthypotenuse=xh\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h}
    • tan⁡θ=oppositeadjacent=yx\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x}
    • Identity: sin⁡θ⋅cos⁡θ= yh⋅xh=yx=tan⁡θ\sin\theta \cdot \cos\theta = \frac{\,y}{h} \cdot \frac{x}{h} = \frac{y}{x} = \tan\theta
    • Reciprocals: csc⁡θ=1sin⁡θ,  sec⁡θ=1cos⁡θ,  cot⁡θ=1tan⁡θ\csc\theta = \frac{1}{\sin\theta}, \; \sec\theta = \frac{1}{\cos\theta}, \; \cot\theta = \frac{1}{\tan\theta}
  • Table 1 (critical angles in the first quadrant) and unit circle coordinates are provided to memorize sine and cosine values at key angles.
  • Technique 9: Problem-solving strategy for right triangles
    • If unknown is a side: use Pythagorean theorem or trig ratios if you know a side and one angle.
    • If unknown is sin, cos, tan of an angle: use the unit-circle coordinates (Table 1) or the definitions, and possibly find needed sides via Pythagoras.
    • Remember radian measure and the reference to critical angle values.
  • Examples (brief):
    • 6.1: Given an angle π/6, find a missing side using sin(π/6) = 1/2, and the value of the hypotenuse.
    • 6.2: Given adjacent and hypotenuse, compute sin θ by finding the opposite via the Pythagorean theorem and then sin θ = opposite/hypotenuse.

Trig Functions

  • Definition: For any real θ, on the unit circle, cos θ is the x-coordinate, sin θ is the y-coordinate, and tan θ = sin θ / cos θ.
  • Critical angles and Table 1 values (sin and cos) provide the necessary coordinates.
  • Technique 10: For a critical angle, sketch the unit circle, locate the angle, read off the coordinates (cos θ, sin θ). Use signs by quadrant to determine sin, cos, tan values.
  • Examples:
    • 7.1: sin(11π/6) = −1/2
    • 7.2: cos(−π/4) = √2/2
    • 7.3: tan(10π/3) = tan(4π/3) = √3
  • Exponents (two types of questions): simplify expressions with exponents; or solve for a variable in the exponent.
  • Exponent rules (E1–E5):
    • (E1) Am⋅An=Am+nA^m \cdot A^n = A^{m+n}
    • (E2) (Am)n=Amn(A^m)^n = A^{mn}
    • (E3) (AB)m=AmBm(A B)^m = A^m B^m
    • (E4) A^{−m} = \frac{1}{A^m}</li><li>(E5)</li> <li>(E5)A^{m/n} = \sqrt[n]{A^m}(assumingn≠0,A≥0ifrealrootsareinvolved)</li></ul></li><li>Importantcaveat:Exponentsdistributeoverproducts,notoversums:(assuming n ≠ 0, A ≥ 0 if real roots are involved)</li></ul></li> <li>Important caveat: Exponents distribute over products, not over sums:(A + B)^m \neq A^m + B^m.Thesameappliestorootsandfractionalexponents.</li><li>Technique11:Tosimplifywithexponents,apply(E1)–(E5)andinterpretfractionalpowersasrootsfirsttoreducemagnitudebeforeapplyingtheexponent.</li><li>Example8.1(caveat):Thetranscript’shandlingof(9x−8)3/2treatstheexpressionasaproductinsidetheparentheses,whichisnotvalid;thecorrectapproachforasum/differenceinsidethebaseistoapplytheexponenttotheentirebaseafterfactoringifpossible,ortoseparateiftheexpressioncanbefactoredintoaproduct.Aprecisesimplificationrequiresaconcretefactorizationof(9x−8)ifpossible;otherwisethisexpressioncannotbesimplifiedfurtherwithoutadditionalcontext.</li><li>Example8.2:(8/125)−2/3→Userule(E5)andreciprocalrule:<ul><li>(8125)−2/3=(1258)2/3=1252/382/3=5222=254</li></ul></li><li>Technique12(solvingexponents):Twomethods<ul><li>(i)Same−basemethod:rewritebothsidesaspowersofthesamebaseandequateexponents.</li><li>(ii)Takelogarithms:applylogtobothsidesandsolvefortheexponent.</li><li>Inmanycases,thesame−basemethodyieldsrationalanswers;logsyieldpotentiallynon−rationalresults.</li></ul></li><li>Example8.3:Solve95x−1=81<ul><li>Since81=92,equateexponents:5x−1=2→x=3/5</li></ul></li><li>Example8.4:Solve3x=16<ul><li>Notacleansame−basecase;uselogs:x=log⁡(16)/log⁡(3)(orx=ln⁡16ln⁡3)orx=log⁡3(16)</li></ul></li></ul><p>Graphing</p><ul><li>Threemaintypestosketch:trigcurves(y=sinx,y=cosx,y=tanx),linear(y=mx+b),andparabola(y=ax2+bx+c).</li><li>Requiredlabeledfeatures:x−intercepts,y−intercept,maxima/minima,asymptotes(fortan),andcoordinatesofverticeswhenapplicable.</li><li>Trigonometricgraphs(Figures5–7inthedocument):sinx,cosx,tanxwiththeirperiodicbehaviorandintercepts.</li><li>Relationship:sinxandcosxarethesameshapeshiftedbyπ/2;sin0=0,cos0=1;sinzerosatmultiplesofπ;coszerosatoddmultiplesofπ/2;tanhasverticalasymptoteswherecosx=0.</li><li>Technique13:Tographstandardtrigfunctions,remembertheshapesandimportantvalues(intercepts,extrema,asymptotes)usingtheunitcirclevaluesfromFigure4.</li><li>Example9.1:Graphy=cosxfor−π≤x≤π,labelinginterceptsandextremepoints.</li><li>Graphinglines(Technique14):Tography=mx+b(notvertical):plotintercepts(0,b)andsolveforx−interceptbysettingy=0;drawlinethroughtwopoints;labelslopem.</li><li>Examplesforlines:<ul><li>9.2:Graphy=2x−4;intercepts(0,−4)and(2,0);slope2.</li><li>9.3:Graphy=−3x;interceptsat(0,0)andanotherpointsuchas(1,−3).</li></ul></li><li>Parabolas(Technique15):Graphy=ax2+bx+cby:<ul><li>Findingintercepts:y−intercept(0,c);x−interceptsbysolvingy=0;vertexx−coordinateis−b/(2a),oruseaverageofx−interceptsiftwoexist;y−coordinateatthevertexisy(−b/(2a)).</li><li>Example9.4:Graphy=x2+4withintercepts(0,4)andnorealx−intercepts;vertexat(0,4).</li><li>Example9.5:Graphy=2x2+x;interceptsat(0,0)andx−interceptswhen2x2+x=0→x(2x+1)=0→x=0orx=−1/2;vertexnearx=−1/4;yatvertexis−1/8.</li></ul></li></ul><p>GeometricMeasurement</p><ul><li>Focus:computegeometricquantitieswithcorrectunits;formulasaregiveninTable2.</li><li>Units:length(perimeter)haslinearunits(cm,in,etc.);areahassquareunits(cm2);volumehascubicunits(cm3).</li><li>Technique16:Picktherightformula,adjustasneeded(one−half,one−third,scalefactors),andensureunitsmatch(squareunitsforarea,cubicunitsforvolume).</li><li>Table2(selected):<ul><li>Perimeter:RectangleP=2(l+w);TriangleP=s1+s2+s3;CircleC=2πr</li><li>Area:RectangleA=lw;CircleA=πr2;TriangleA=1/2bh</li><li>Volume:BoxV=lwh;SphereV=(4/3)πr3;ConeV=(1/3)πr2h;CylinderV=πr2h</li></ul></li><li>Examples:<ul><li>Example10.1:Perimeteroftrianglewithsides5cm,6cm,1cm→P=5+6+1=12cm</li><li>Example10.2:Volumeofspherewithradius3ft→V=(4/3)π(3)3=36πft3</li><li>Example10.3:Areaofacirclesectorwithradius7ftandangleπ/3radians</li><li>Fractionofcircle:θ/(2π)=(π/3)/(2π)=1/6</li><li>Area:Asector=(1/6)⋅(πr2)=(1/6)⋅(π⋅49)=49π/6ft2</li></ul></li></ul><p>Notesandclarifications</p><ul><li>Theprovidedtranscriptcontainsseveraltypographicalerrorsinsomenumericalsteps(e.g.,slopesignsinExample5.2,incorrectarithmeticinseveralplaces).Thenotesaboveprovidethestandard,correctresolutionsforthoseproblems.Whenpracticing,verifyresultsbyexpandingproductstocheckfactoringorbysubstitutingbackintooriginalequations.</li><li>AllformulasarepresentedinLaTeX−compatibleformatforyourstudynotes(enclosedin. The same applies to roots and fractional exponents.</li> <li>Technique 11: To simplify with exponents, apply (E1)–(E5) and interpret fractional powers as roots first to reduce magnitude before applying the exponent.</li> <li>Example 8.1 (caveat): The transcript’s handling of (9x − 8)^{3/2} treats the expression as a product inside the parentheses, which is not valid; the correct approach for a sum/difference inside the base is to apply the exponent to the entire base after factoring if possible, or to separate if the expression can be factored into a product. A precise simplification requires a concrete factorization of (9x − 8) if possible; otherwise this expression cannot be simplified further without additional context.</li> <li>Example 8.2: (8/125)^{−2/3} → Use rule (E5) and reciprocal rule:<ul> <li>\left(\frac{8}{125}\right)^{−2/3} = \left(\frac{125}{8}\right)^{2/3} = \frac{125^{2/3}}{8^{2/3}} = \frac{5^{2}}{2^{2}} = \frac{25}{4}</li></ul></li> <li>Technique 12 (solving exponents): Two methods<ul> <li>(i) Same-base method: rewrite both sides as powers of the same base and equate exponents.</li> <li>(ii) Take logarithms: apply log to both sides and solve for the exponent.</li> <li>In many cases, the same-base method yields rational answers; logs yield potentially non-rational results.</li></ul></li> <li>Example 8.3: Solve 9^{5x−1} = 81<ul> <li>Since 81 = 9^2, equate exponents: 5x − 1 = 2 → x = 3/5</li></ul></li> <li>Example 8.4: Solve 3^x = 16<ul> <li>Not a clean same-base case; use logs: x = \log(16)/\log(3) (or x = \frac{\ln 16}{\ln 3}) or x = \log_3(16)</li></ul></li> </ul> <p>Graphing</p> <ul> <li>Three main types to sketch: trig curves (y = sin x, y = cos x, y = tan x), linear (y = mx + b), and parabola (y = ax^2 + bx + c).</li> <li>Required labeled features: x-intercepts, y-intercept, maxima/minima, asymptotes (for tan), and coordinates of vertices when applicable.</li> <li>Trigonometric graphs (Figures 5–7 in the document): sin x, cos x, tan x with their periodic behavior and intercepts.</li> <li>Relationship: sin x and cos x are the same shape shifted by π/2; sin 0 = 0, cos 0 = 1; sin zeros at multiples of π; cos zeros at odd multiples of π/2; tan has vertical asymptotes where cos x = 0.</li> <li>Technique 13: To graph standard trig functions, remember the shapes and important values (intercepts, extrema, asymptotes) using the unit circle values from Figure 4.</li> <li>Example 9.1: Graph y = cos x for −π ≤ x ≤ π, labeling intercepts and extreme points.</li> <li>Graphing lines (Technique 14): To graph y = mx + b (not vertical): plot intercepts (0, b) and solve for x-intercept by setting y = 0; draw line through two points; label slope m.</li> <li>Examples for lines:<ul> <li>9.2: Graph y = 2x − 4; intercepts (0, −4) and (2, 0); slope 2.</li> <li>9.3: Graph y = −3x; intercepts at (0, 0) and another point such as (1, −3).</li></ul></li> <li>Parabolas (Technique 15): Graph y = ax^2 + bx + c by:<ul> <li>Finding intercepts: y-intercept (0, c); x-intercepts by solving y = 0; vertex x-coordinate is −b/(2a), or use average of x-intercepts if two exist; y-coordinate at the vertex is y(−b/(2a)).</li> <li>Example 9.4: Graph y = x^2 + 4 with intercepts (0, 4) and no real x-intercepts; vertex at (0, 4).</li> <li>Example 9.5: Graph y = 2x^2 + x; intercepts at (0, 0) and x-intercepts when 2x^2 + x = 0 → x(2x + 1) = 0 → x = 0 or x = −1/2; vertex near x = −1/4; y at vertex is −1/8.</li></ul></li> </ul> <p>Geometric Measurement</p> <ul> <li>Focus: compute geometric quantities with correct units; formulas are given in Table 2.</li> <li>Units: length (perimeter) has linear units (cm, in, etc.); area has square units (cm^2); volume has cubic units (cm^3).</li> <li>Technique 16: Pick the right formula, adjust as needed (one-half, one-third, scale factors), and ensure units match (square units for area, cubic units for volume).</li> <li>Table 2 (selected):<ul> <li>Perimeter: Rectangle P = 2(l + w); Triangle P = s1 + s2 + s3; Circle C = 2πr</li> <li>Area: Rectangle A = l w; Circle A = πr^2; Triangle A = 1/2 bh</li> <li>Volume: Box V = l w h; Sphere V = (4/3) π r^3; Cone V = (1/3) π r^2 h; Cylinder V = π r^2 h</li></ul></li> <li>Examples:<ul> <li>Example 10.1: Perimeter of triangle with sides 5 cm, 6 cm, 1 cm → P = 5 + 6 + 1 = 12 cm</li> <li>Example 10.2: Volume of sphere with radius 3 ft → V = (4/3) π (3)^3 = 36π ft^3</li> <li>Example 10.3: Area of a circle sector with radius 7 ft and angle π/3 radians</li> <li>Fraction of circle: θ/(2π) = (π/3)/(2π) = 1/6</li> <li>Area: A_sector = (1/6)·(π r^2) = (1/6)·(π·49) = 49π/6 ft^2</li></ul></li> </ul> <p>Notes and clarifications</p> <ul> <li>The provided transcript contains several typographical errors in some numerical steps (e.g., slope signs in Example 5.2, incorrect arithmetic in several places). The notes above provide the standard, correct resolutions for those problems. When practicing, verify results by expanding products to check factoring or by substituting back into original equations.</li> <li>All formulas are presented in LaTeX-compatible format for your study notes (enclosed in…$$ when displayed).
    • The ten topics form a compact review of essential concepts needed for the ABC Calculus I exam, including algebra, trigonometry, exponents, geometry, and graphing. Focus on understanding the techniques, not just memorizing results, and pay attention to domain restrictions (denominators, zeros) when solving equations and inequalities.