CLEP Precalculus Quick Review

Every polynomial simplification and factoring problem is really about rewriting an expression in an equivalent form to reveal its roots or simplify computation.

Difference of squares $a^2-b^2=(a-b)(a+b)$; sum/difference of cubes $a^3± b^3=(a± b)(a^2 ab+b^2)$; perfect square trinomial $(a± b)^2=a^2±2ab+b^2$; factor by grouping for four-term expressions.

$(a+b)^2$ (expands to three terms, $a^2+2ab+b^2$) vs.\ $a^2+b^2$ (only two terms --- these are never equal in general).

Forgetting the middle term when squaring a binomial, or mis-distributing a negative sign across a subtraction inside parentheses.

Square a binomial $arrow$ 3 terms, not 2.

Linear equations isolate $x$ using inverse operations; inequalities behave the same way except the direction flips when multiplying or dividing by a negative.

$|x-a|<b ⇒ a-b<x<a+b$ (for $b>0$); $|x-a|>b ⇒ x<a-b$ or $x>a+b$; compound inequalities are solved by isolating $x$ in the middle of the chain.

An absolute-value ``less than'' statement (gives one connected interval, an AND) vs.\ a ``greater than'' statement (gives two separate rays, an OR).

Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number.

Abs.\ value $<$ $arrow$ between; abs.\ value $>$ $arrow$ split.

Every quadratic can be solved by factoring, completing the square, or the quadratic formula, and the discriminant predicts the nature of the roots before you even solve.

$x=/-b±sqrtb^2-4ac2a$; discriminant $=b^2-4ac$: $>0$ gives two real roots, $=0$ gives one repeated real root, $<0$ gives two complex conjugate roots.

The vertex $x$-coordinate $x=-/b2a$ vs.\ the full quadratic formula, which adds $±/sqrtb^2-4ac2a$ to that same base value.

Dropping the $±$ symbol, or mis-simplifying $sqrtb^2-4ac$ when $b$ itself is negative (sign errors before squaring).

$<0 arrow$ no real solutions; graph never touches the $x$-axis.

Higher-degree polynomial equations are solved by finding rational roots, factoring them out, and reducing to a degree you can already solve.

Rational Root Theorem: candidate roots $=±/factors of constant termfactors of leading coefficient$; use synthetic division to test and reduce; Fundamental Theorem of Algebra --- a degree-$n$ polynomial has exactly $n$ roots counting multiplicity and complex roots.

A root's multiplicity (how many times a factor repeats) vs.\ the total count of distinct roots on the graph.

Stopping after finding one real root instead of fully reducing the polynomial to locate all $n$ roots.

Degree $n$ $arrow$ exactly $n$ roots, counted with multiplicity.

Rational equations are solved by clearing denominators, but every proposed solution must be checked against excluded values; rational inequalities need a full sign chart, not cross-multiplication.

Multiply both sides by the LCD to clear denominators; for inequalities, move everything to one side, find critical numbers (zeros and undefined points), and test the sign of the expression in each interval.

An extraneous solution (algebraically valid but makes a denominator zero) vs.\ a solution that is simply outside the equation's domain from the start.

Multiplying a rational inequality by a variable expression without knowing its sign, which silently flips or invalidates the inequality.

Rational inequality $arrow$ sign chart, never cross-multiply blindly.

Solving a radical equation means isolating the radical and raising both sides to a matching power, a step that can introduce extraneous solutions.

Isolate the radical first, then raise both sides to the index's power (square for $sqrt\ $, cube for $sqrt[3]\ $); always substitute every solution back into the original equation to check it.

Squaring both sides of an equation (a valid, reversible-with-checking technique) vs.\ squaring both sides of an inequality (not generally valid).

Forgetting to check for extraneous roots after squaring --- a ``solution'' that makes the original radical expression negative or undefined.

Squared both sides? $arrow$ ALWAYS check for extraneous roots.

Exponential and logarithmic equations are inverses of one another, so you convert between the two forms to isolate the variable; systems combine multiple equations to find a shared solution.

$b^x=y _b y=x$; to solve $_b(x)=_b(y)$, set $x=y$ (subject to the domain); systems of equations are solved by substitution, elimination, or graphing for the intersection point(s).

$(xy)= x+ y$ (log of a product) vs.\ $(x+y)$, which does not simplify into separate logs.

Taking $$ of both sides but forgetting the domain restriction $x>0$, then accepting an extraneous negative ``solution.''

$_b y=x arrow$ rewrite as $b^x=y$ to unlock it.

A function assigns exactly one output to every input; domain and range describe which inputs and outputs are actually valid.

Vertical Line Test: a graph represents a function iff no vertical line crosses it more than once; domain restrictions come from denominators ($0$), even roots (radicand $≥0$), and logarithms (argument $>0$).

Domain (the set of valid $x$-values that can go in) vs.\ range (the set of resulting $y$-values that come out).

Forgetting to exclude a denominator's zero, or forgetting a negative-value restriction under an even radical, when stating the domain.

Function $arrow$ one output per input; vertical line test confirms it.

Function notation $f(x)$ is a substitution instruction --- replace every $x$ in the rule with the given input.

$f(a)$ means substitute $x=a$ into $f$'s rule; difference quotient $/f(x+h)-f(x)h$; a piecewise function is evaluated by first checking which branch's condition the input satisfies.

$f(x+h)$ (substitute the whole expression $x+h$ everywhere $x$ appears) vs.\ $f(x)+h$ (evaluate $f$ first, then add $h$ afterward).

Distributing incorrectly into $f(x+h)$, e.g., writing $f(x)+f(h)$ for a non-linear function $f$.

$f(x+h) f(x)+f(h)$ unless $f$ is linear.

Functions combine algebraically (add, subtract, multiply, divide) or by composition, where the output of one function becomes the input of another.

$(f g)(x)=f(g(x))$; the domain of $f g$ requires $x$ in the domain of $g$ AND $g(x)$ in the domain of $f$; $(f/g)(x)$ additionally requires $g(x)0$.

$(f g)(x)$ vs.\ $(g f)(x)$ --- function composition is generally not commutative.

Computing $f(g(x))$ by multiplying $f(x)· g(x)$ instead of substituting $g(x)$ in for every $x$ in $f$.

$f g$: work inside out --- evaluate $g$ first.

An inverse function undoes exactly what the original function does, and it exists only where the original function is one-to-one.

$f^-1(f(x))=x$ and $f(f^-1(x))=x$; to find $f^-1$: swap $x$ and $y$, then solve for $y$; the graphs of $f$ and $f^-1$ are reflections of each other across $y=x$; a function has an inverse iff it passes the Horizontal Line Test.

$f^-1(x)$ (the inverse function) vs.\ $/1f(x)=[f(x)]^-1$ (the reciprocal) --- these are almost always different functions.

Writing $f^-1(x)=1/f(x)$ purely from the $-1$ exponent notation, without actually inverting the function.

Inverse $arrow$ swap $x,y$, solve for $y$; reflects over $y=x$.

A function's symmetry --- even, odd, or neither --- is tested algebraically with $f(-x)$ and predicts symmetry in its graph.

Even: $f(-x)=f(x)$, graph symmetric about the $y$-axis (e.g., $x^2$, $ x$); Odd: $f(-x)=-f(x)$, graph symmetric about the origin (e.g., $x^3$, $ x$).

An ``odd function'' (a symmetry property tested with $f(-x)$) vs.\ an ``odd-degree polynomial'' --- an odd-degree polynomial is not automatically an odd function unless every term has an odd exponent.

Assuming a function is even or odd purely from its degree, without actually testing $f(-x)$ algebraically.

$f(-x)=f(x)arrow$ even; $f(-x)=-f(x)arrow$ odd.

A linear function has a constant rate of change, so its graph is a straight line fully determined by a slope and one point.

Slope $m=/y_2-y_1x_2-x_1$; slope-intercept form $y=mx+b$; point-slope form $y-y_1=m(x-x_1)$; parallel lines share equal slopes, perpendicular lines have slopes that are negative reciprocals ($m_1 m_2=-1$).

Point-slope form (built directly from a point and a slope) vs.\ slope-intercept form (isolated for $y$, revealing the $y$-intercept directly).

Subtracting coordinates in mismatched order when computing $ y/ x$ from two points.

Perpendicular slopes multiply to $-1$.

A quadratic function's graph is a parabola whose vertex, direction, and width are read directly off its vertex form.

Vertex form $f(x)=a(x-h)^2+k$, vertex $(h,k)$; standard form $f(x)=ax^2+bx+c$, vertex at $x=-/b2a$; $a>0$ opens up (minimum), $a<0$ opens down (maximum); larger $|a|$ makes the parabola narrower.

Vertex form's $(x-h)$: the vertex $x$-coordinate is $+h$, not $-h$, even though the formula shows subtraction.

Reading the vertex as $(-h,k)$ instead of $(h,k)$ because of the minus sign inside the parentheses.

$(x-h)^2+k arrow$ vertex is $(h,k)$ --- the sign flips!

A polynomial's degree and leading coefficient determine its end behavior, while its factored form reveals its zeros and each zero's multiplicity.

Leading term test: even degree with positive leading coefficient $arrow$ both ends rise; odd degree with positive leading coefficient $arrow$ falls left, rises right; a zero of even multiplicity touches and turns at the $x$-axis, a zero of odd multiplicity crosses straight through.

A zero with multiplicity 2 (graph touches and bounces back) vs.\ a zero with multiplicity 1 or 3 (graph crosses straight through).

Assuming every zero crosses the $x$-axis and missing that even-multiplicity zeros only touch and bounce.

Even multiplicity $arrow$ bounce; odd multiplicity $arrow$ cross.

A rational function's graph is shaped by vertical asymptotes (from the denominator's zeros) and horizontal or slant asymptotes (from comparing the numerator's and denominator's degrees).

Vertical asymptote at $x=a$ where the denominator $=0$ and the numerator $0$ (a ``hole'' occurs instead if a common factor cancels); if $(num)<(denom)$, horizontal asymptote $y=0$; if degrees are equal, $y=$ ratio of leading coefficients; if $(num)=(denom)+1$, there is a slant asymptote found by polynomial division.

A removable discontinuity (a ``hole,'' from a common factor that cancels) vs.\ a true vertical asymptote (a denominator zero that does NOT cancel).

Forgetting to check for a cancelling common factor first, mistaking what is actually a hole for a vertical asymptote.

Degrees equal $arrow$ horizontal asymptote = ratio of leading coefficients.

A piecewise function is graphed and evaluated one branch at a time, using the stated domain restriction to select the correct rule.

Check which interval the input falls into before applying the corresponding rule; strict inequalities ($<,>$) give an open circle at that boundary, non-strict inequalities ($≤,≥$) give a closed/filled circle.

An open circle (boundary value excluded) vs.\ a closed/filled circle (boundary value included) on a piecewise graph.

Evaluating a piecewise function at a boundary point using the wrong branch, or graphing both branches with filled circles at a shared boundary and accidentally creating a non-function.

Match the input to its interval FIRST, then plug in.

Exponential functions model constant-percentage growth or decay; the base determines whether the graph rises or falls, and the graph always has a horizontal asymptote.

$f(x)=ab^x$; growth if $b>1$, decay if $0<b<1$; horizontal asymptote $y=0$ (unless shifted); $y$-intercept $=a$; domain is all reals, range is $y>0$ (before vertical shifting).

Exponential decay ($0<b<1$) vs.\ a growth base with a negative exponent ($b^-x=(1/b)^x$) --- algebraically these produce the same decaying shape.

Confusing $-b^x$ (reflects the graph over the $x$-axis, output becomes negative) with $b^-x$ (reflects over the $y$-axis, output stays positive).

$b>1 arrow$ growth; $0<b<1 arrow$ decay; asymptote always $y=0$.

A logarithmic function is the inverse of an exponential function, so its graph is the exponential's reflection over $y=x$, complete with a vertical asymptote.

$y=_b x b^y=x$; domain $x>0$, range all reals; vertical asymptote $x=0$ (unless shifted); $_b(xy)=_b x+_b y$; $_b(x/y)=_b x-_b y$; $_b(x^k)=k_b x$; change of base $_b x=/ x b$.

$(x)·(y)$ (a product of two separate logs) vs.\ $(xy)$ (the log of a product) --- these are NOT equal.

``Simplifying'' $(x+y)$ as $ x+ y$ --- the sum/difference rule only applies to logs of products/quotients, never to logs of a sum.

$_b(xy)=_b x+_b y$, but $_b(x+y)$ never simplifies.

Every graph transformation follows a predictable rule based on where the constant appears relative to $x$, inside or outside the function.

$f(x)+k$ shifts up $k$; $f(x)-k$ shifts down; $f(x-h)$ shifts right $h$; $f(x+h)$ shifts left $h$; $-f(x)$ reflects over the $x$-axis; $f(-x)$ reflects over the $y$-axis; $af(x)$ stretches vertically if $|a|>1$, compresses if $0<|a|<1$; $f(bx)$ compresses horizontally if $|b|>1$.

$f(x-h)$, which shifts RIGHT (opposite of the sign inside), vs.\ $f(x)+k$, which shifts UP (same sign as outside).

Shifting $f(x+3)$ to the right instead of the left --- the sign inside the parentheses gives the opposite of the shift direction.

Inside the parentheses $arrow$ opposite direction, horizontal; outside $arrow$ same direction, vertical.

The same function can be represented as an equation, a graph, or a table of values, and CLEP frequently asks you to translate between the three.

Constant first differences between consecutive table outputs $arrow$ linear function; constant ratios between consecutive outputs $arrow$ exponential function; constant second differences $arrow$ quadratic function.

Constant differences in a table (signals a linear function) vs.\ constant ratios in a table (signals an exponential function).

Misreading a table with constant ratios as linear because the differences merely look ``roughly steady'' instead of checking the ratio precisely.

Table: differences constant $arrow$ linear; ratios constant $arrow$ exponential.

Reading a graph for where a function increases, decreases, or reaches a max/min reveals its behavior without needing calculus.

A function is increasing on an interval where $f(x)$ rises as $x$ increases; a local max/min occurs where the function switches from increasing to decreasing (or vice versa); end behavior describes what $f(x)$ does as $x→±∞$.

A local (relative) maximum/minimum (the highest/lowest point in a neighborhood) vs.\ the global (absolute) maximum/minimum (the highest/lowest value over the entire domain).

Reporting a local max as though it were the function's absolute max without checking the rest of the domain or its end behavior.

Increasing/decreasing describes $y$ as $x$ moves left to right.

Analytic geometry translates geometric relationships --- like distance and the halfway point between two locations --- into algebraic formulas built from coordinates.

Distance $d=sqrt(x_2-x_1)^2+(y_2-y_1)^2$; midpoint $M=≤ft(/x_1+x_22,/y_1+y_22)$.

The distance formula (produces a single number) vs.\ the midpoint formula (produces a point, an ordered pair).

Forgetting to divide by 2 in the midpoint formula, or forgetting the square root in the distance formula.

Distance $arrow$ a number; midpoint $arrow$ a point.

A circle is the set of all points equidistant from a fixed center, and completing the square converts a circle's general equation into a form that reveals the center and radius directly.

Standard form $(x-h)^2+(y-k)^2=r^2$, center $(h,k)$, radius $r$; convert from general form $x^2+y^2+Dx+Ey+F=0$ by completing the square in both $x$ and $y$.

The center $(h,k)$, read with the same sign flip as vertex form, vs.\ the radius, which is $sqrtr^2$, not the right-hand-side constant $r^2$ itself.

Reporting the radius as $r^2$ (the equation's right-hand constant) instead of taking its square root.

$(x-h)^2+(y-k)^2=r^2 arrow$ center $(h,k)$, radius $=sqrtr^2$.

Geometrically, a parabola is the set of points equidistant from a fixed focus and a fixed directrix --- a conic description that goes beyond the algebra-class vertex form.

Vertical-axis parabola $(x-h)^2=4p(y-k)$, vertex $(h,k)$, focus $(h,k+p)$, directrix $y=k-p$; horizontal-axis parabola $(y-k)^2=4p(x-h)$, focus $(h+p,k)$, directrix $x=h-p$.

The algebra-class vertex form $y=a(x-h)^2+k$ vs.\ the conic form $(x-h)^2=4p(y-k)$ used to locate the focus and directrix.

Mixing up which variable is squared, which determines whether the parabola opens vertically or horizontally.

Squared variable $arrow$ that axis is the parabola's axis of symmetry.

Ellipses and hyperbolas are each defined by a sum or difference of distances to two foci, and their standard-form equations reveal orientation and shape straight from the denominators.

Ellipse $/(x-h)^2a^2+/(y-k)^2b^2=1$ (major axis lies along the larger denominator), with $c^2=a^2-b^2$; hyperbola $/(x-h)^2a^2-/(y-k)^2b^2=1$ (opens left--right) or $/(y-k)^2a^2-/(x-h)^2b^2=1$ (opens up--down), with $c^2=a^2+b^2$.

$c^2=a^2-b^2$ for an ellipse vs.\ $c^2=a^2+b^2$ for a hyperbola --- the sign flips between the two conics.

Assuming the larger denominator is always under the $x$-term (horizontal axis) without checking which variable it actually sits under.

Ellipse: a plus sign between terms; hyperbola: a minus sign between terms.

In a right triangle, the trig ratios relate an acute angle to the ratio of two specific sides, giving a way to find a missing side or angle.

$=/opphyp$, $=/adjhyp$, $=/oppadj$; Pythagorean theorem $a^2+b^2=c^2$; mnemonic SOH-CAH-TOA.

``Opposite'' and ``adjacent'' are defined relative to whichever acute angle is currently being used, not fixed to specific sides of the triangle.

Mislabeling opposite vs.\ adjacent when the reference angle switches to the triangle's other acute angle.

SOH-CAH-TOA --- always relative to the angle in question.

The unit circle links angles directly to exact coordinates, giving exact trig values at key angles without needing a calculator.

Radians $=/180×$ degrees; on the unit circle, the point at angle $$ is $(,)$; know exact values at $0,/6,/4,/3,/2$ and their reference-angle counterparts in every quadrant.

Degree mode vs.\ radian mode --- mixing them up produces a wildly wrong numeric answer even when the setup is otherwise correct.

Working in the wrong angle mode (degrees vs.\ radians), especially on Section 2, where exact-value reasoning must be done without a calculator to catch the error.

$180^=$ radians $arrow$ convert by multiplying by $/180$ or $180/$.

Sine and cosine graphs are periodic waves whose amplitude, period, phase shift, and vertical shift are all read directly from the equation.

$y=A(B(x-C))+D$: amplitude $=|A|$, period $=/2B$, phase shift $=C$, vertical shift $=D$, midline $y=D$.

The period formula $2/B$ (uses $B$, the coefficient of $x$) vs.\ the amplitude (uses $A$, the coefficient out front).

Computing the period as $2· B$ instead of $2/B$, or forgetting to factor $B$ out of $(Bx-C)$ before reading off the phase shift.

Period $=2/B$; amplitude $=|A|$.

Tangent, cotangent, secant, and cosecant graphs behave differently from sine/cosine --- they have vertical asymptotes rather than a bounded amplitude.

$ x$ has period $$, with vertical asymptotes where $ x=0$ (i.e., $x=/2+n$); $ x=1/ x$ has asymptotes where $ x=0$; $ x=1/ x$ has asymptotes where $ x=0$.

The period of tangent ($$) vs.\ the period of sine/cosine ($2$) --- tangent repeats twice as fast.

Applying the sine/cosine period formula ($2/B$) directly to tangent, or assuming tangent has an amplitude the way sine and cosine do.

Tan/cot period $=$; sin/cos period $=2$; sec/csc mirror cos/sin's asymptote locations.

Trig identities are algebraic tools for rewriting an expression into an equivalent, more useful form; most CLEP identity questions reduce to a Pythagorean or sum/difference identity.

Pythagorean identities: $^2+^2=1$, $1+^2=^2$, $1+^2=^2$; double angle: $(2)=2$, $(2)=^2-^2$; sum formulas: $(a± b)= a b± a b$, $(a± b)= a b a b$.

$(2)=^2-^2$ (a correct double-angle form) vs.\ the invalid shortcut $(2)=2$.

Distributing trig functions like multiplication, e.g., writing $(a+b)= a+ b$ instead of applying the actual sum formula.

$^2+^2=1$ is the master identity --- everything else builds from it.

Inverse trig functions return the one angle, in a restricted range, whose trig ratio matches the input; solving a trig equation requires accounting for the periodic, multi-valued nature of trig functions.

$ x$ has range $[-/2,/2]$; $ x$ has range $[0,]$; $ x$ has range $(-/2,/2)$; general solutions to a trig equation add $2n$ (or $n$ for tangent) to capture every coterminal/periodic answer.

The single restricted output of an inverse trig function vs.\ the full, infinite solution set of a trig equation on a stated interval.

Giving only one solution to $ x=k$ on an interval and missing the second solution in the other quadrant that shares the same reference angle.

Inverse trig $arrow$ one restricted answer; ``solve the equation'' $arrow$ find ALL answers in the interval.

For non-right triangles, the Law of Sines and Law of Cosines extend trigonometry beyond SOH-CAH-TOA to solve for missing sides and angles.

Law of Sines $/a A=/b B=/c C$ (use with an angle--opposite-side pair); Law of Cosines $c^2=a^2+b^2-2ab C$ (use with SAS or SSS).

The Law of Sines' ambiguous case (SSA can produce 0, 1, or 2 valid triangles) vs.\ the Law of Cosines (SAS/SSS always determines a unique triangle).

Applying the Law of Sines to an SSA situation without checking the ambiguous case, and missing a second valid triangle.

Know an angle & its opposite side $arrow$ Law of Sines; know SAS/SSS $arrow$ Law of Cosines.

Real-world data with a constant rate of change, a single max/min, or multiple turning points is modeled with a linear, quadratic, or higher-degree polynomial function, respectively.

Constant rate of change $arrow$ linear model; a single maximum or minimum (e.g., projectile height) $arrow$ quadratic model $h(t)=-/12 gt^2+v_0t+h_0$; multiple turning points $arrow$ higher-degree polynomial model.

A ``maximum value'' question (asking for the $y$-value at the vertex) vs.\ a ``when does it reach the max'' question (asking for the $x$-value/time at the vertex).

Answering a max-height question with the time value (or vice versa) instead of evaluating the function at the vertex to get the actual height.

Vertex $x=-b/2a$ gives WHEN; $f$ at that $x$ gives the max/min VALUE.

Quantities that grow or decay by a constant percentage --- population, compound interest, radioactive decay --- are modeled with exponential functions; logarithms invert those models to solve for time.

$A=A_0(1+r)^t$ (growth/decay by rate $r$); compound interest $A=P≤ft(1+/rn)^nt$; continuous growth $A=A_0e^rt$; solve for $t$ by taking a logarithm of both sides.

The growth/decay factor $b=(1+r)$ vs.\ the rate $r$ itself --- it's easy to report $b$ when the question asks for the percentage rate, or vice versa.

Using $r$ directly as the base instead of $(1+r)$, or dividing instead of taking a logarithm when solving for a variable in the exponent.

``Solve for the exponent'' $arrow$ take a $$ of both sides.

Periodic real-world phenomena --- tides, temperature cycles, circular motion --- are modeled with sine or cosine functions whose parameters map directly onto the physical situation.

Amplitude = half the distance between the max and min values; period = time for one full cycle; midline $D=/max+min2$; general model $y=A≤ft(/2period(x-C))+D$.

The amplitude (half the total swing) vs.\ the full range (max $-$ min), which is twice the amplitude.

Setting the amplitude equal to the full max-to-min distance instead of half of it.

Amplitude $=/max-min2$; midline $=/max+min2$.

Eleven formulas cover nearly every computation box on this exam --- know every symbol cold.

When two terms sound alike on this exam, assume the question is testing the difference.

Match the data pattern to the family first --- the graph shape and domain/range follow automatically.

Learn Quadrant I cold, then apply the ASTC sign rule to every other quadrant.

Domain $arrow$ intercepts $arrow$ symmetry $arrow$ asymptotes $arrow$ end behavior $arrow$ plot.

$$48 questions, 90 minutes, two sections --- calculator in Section 1 only, none in Section 2.

Factor $arrow$ cancel (hole) $arrow$ asymptotes $arrow$ intercepts $arrow$ sign chart $arrow$ sketch.

Pattern of change identifies the family; matching given values to parameters builds the equation.

Every ``always/never'' trap statement above is FALSE --- CLEP loves testing the exception.