Concordia University · MATH 201

Concordia MATH 201: Elementary Functions tutoring

Work with an online tutor on functions, graphs, logarithms and trigonometry in Concordia MATH 201. Start with a function or equation, even if you do not know how to begin. Your tutor can explain composition, inverse functions or domain restrictions and guide you through a related question.

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Online 60 min: $55 CAD · Pricing Midterm & final exam help Exam review with explained answers Try a free practice question Review algebra before calculus

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Course overview

Concordia MATH 201 studies power, polynomial, rational, algebraic, exponential, logarithmic, and trigonometric functions. It emphasizes graphs, transformations, inverses, composition, and trigonometric relationships.

Start here

Rational graphs: holes, asymptotes and intercepts
Follow a function composition and its domain
Try the four-question functions review
See how tutoring can help with functions

Rational graphs without calculus: keep the missing inputs

Use factoring, signs and division to describe a rational graph. You do not need derivatives. A rational function is a quotient of polynomials, with every zero of its original denominator excluded. Cancelling a common factor simplifies the values at allowed inputs; it does not restore a missing input.

Choose your scope: the Fall 2026 MATH 201 outline places rational functions in week 4 and the midterm on weeks 1–6. For that early scope, start with domain restrictions, functions and inverses, rational graphs below or logarithms. Trigonometry starts in week 7: keep the trigonometric equation check and triangle practice for later-course preparation unless your instructor announces otherwise. Follow your section’s current instructions; this workshop is original practice, not an official or complete exam.

A five-step graph plan

Factor and record exclusions from the original denominator.
Cancel common factors only where the original expression is defined. A cancelled factor gives a hole only when the remaining expression is finite there. A remaining denominator zero can still give a vertical asymptote.
For an x-intercept, solve for a zero output and check the input is allowed. For a y-intercept, evaluate at zero only if zero is in the domain.
Use polynomial division, or compare degrees, for end behaviour. A horizontal asymptote describes large positive and negative inputs; it is not a line the graph is forbidden to cross.
Check signs on either side of each vertical asymptote. Mark holes with open circles, asymptotes with dashed lines and actual intercepts with filled points. These features guide a sketch; they do not by themselves prove every turning point.

Worked example 1: one hole and one vertical asymptote

Describe $f(x)=(x^2-x-2)/(x^2-4)$ : domain, missing point, asymptotes and intercepts. Sketch the branches without differentiating.

First hint

Factor both quadratics. Keep both original exclusions before cancelling anything.

Follow the complete graph investigation

Factor: $f(x)=\frac{(x-2)(x+1)}{(x-2)(x+2)}=\frac{x+1}{x+2},\qquad x\ne-2,2.$ The domain is all real numbers except $-2$ and $2$ .
At $x=2$ , the simplified expression is finite: $3/4$ . The original function is still undefined there, so mark a hole at $(2,3/4)$ , not a zero or a vertical asymptote.
At $x=-2$ , the remaining denominator vanishes but the numerator is $-1$ . There is a vertical asymptote $x=-2$ . Writing $f(x)=1-1/(x+2)$ shows that values grow without bound positively as $x$ approaches $-2$ from the left, and negatively from the right.
The only x-intercept is $(-1,0)$ : the cancelled zero $x=2$ was excluded. Zero is allowed, giving y-intercept $(0,1/2)$ .
The remainder $-1/(x+2)$ approaches zero for large $|x|$ , so the horizontal asymptote is $y=1$ . The left branch lies above 1, and the right branch below 1, with an open circle at $(2,3/4)$ . The graph never equals 1 because $-1/(x+2)$ is never zero.

Check a nearby value: $f(3)=4/5$ , a point below the horizontal asymptote. A sketching tool might join across a tiny hole, so retain the algebraic domain even when the hole is hard to see.

Worked example 2: an oblique asymptote from division

Find the key graph features of $g(x)=(x^2+1)/(x-1)$ . Explain the oblique (slant) asymptote without using derivatives.

First hint

Multiply $(x-1)(x+1)$ . What must be added to obtain $x^2+1$ ?

Follow the division and graph checks

The only excluded input is $x=1$ . Since $x^2+1>0$ for every real input, there is no common factor and no hole.
Division gives $g(x)=x+1+\frac{2}{x-1}.$ The difference $g(x)-(x+1)=2/(x-1)$ approaches zero for large positive or negative $x$ . Thus $y=x+1$ is the oblique asymptote. There is no horizontal asymptote. The difference never vanishes, so this particular graph never meets its oblique asymptote.
The vertical asymptote is $x=1$ . Near 1 the numerator stays positive: approaching from the left gives large negative outputs, and from the right gives large positive outputs.
There is no real x-intercept because $x^2+1=0$ has no real solution. The y-intercept is $(0,-1)$ .
For $x<1$ , the remainder $2/(x-1)$ is negative, so the graph lies below its oblique asymptote; for $x>1$ , it lies above. Useful checkpoints are $g(-1)=-1$ and $g(2)=5$ .

Sketch honestly: show these verified features without claiming an exact location for a turning point. An asymptote is about the gap between the curve and a line, not about the output approaching a finite number when that line is oblique.

Four original problems: predict, sketch, then check

1. A repeated denominator factor

For $p(x)=(x+1)/(x-2)^2$ , state the domain, holes, intercepts and asymptotes. Do the two sides of the vertical asymptote go in opposite directions?

Hint 1

Close to 2, the numerator is positive and the squared denominator is positive on both sides.

Explained solution 1

The domain excludes 2. Nothing cancels, so there is no hole. The x-intercept is $(-1,0)$ , and the y-intercept is $(0,1/4)$ . The vertical asymptote is $x=2$ ; on both sides the outputs grow without bound positively. The numerator has smaller degree than the denominator, giving horizontal asymptote $y=0$ . A repeated denominator factor does not force opposite-sided behaviour. The graph crosses its horizontal asymptote at $(-1,0)$ .

2. Cancellation does not restore the y-intercept

Investigate $q(x)=x(x-3)/[x(x+2)]$ . Find the domain, hole, intercepts and asymptotes. Explain why substituting zero after cancellation would mislead you.

Hint 2

Record both denominator zeros first; then use $(x-3)/(x+2)$ only on the original domain.

Explained solution 2

The domain excludes $0$ and $-2$ . Cancellation gives $(x-3)/(x+2)=1-5/(x+2)$ on that domain. There is a hole at $(0,-3/2)$ , not a y-intercept. The x-intercept is $(3,0)$ , the vertical asymptote is $x=-2$ , and the horizontal asymptote is $y=1$ . At $-2$ , outputs grow positively from the left and negatively from the right. The original denominator is zero at zero, so the original graph has no point on the y-axis.

3. Cross a horizontal asymptote

For $u(x)=(x^2+x)/(x^2+1)$ , find the domain, both types of axis intercept and any vertical or horizontal asymptote. Find where the graph crosses its horizontal asymptote.

Hint 3

The denominator is always positive. Write $u(x)=1+(x-1)/(x^2+1)$ and ask when the remainder is zero.

Explained solution 3

The domain is all real numbers; there are no holes or vertical asymptotes. Factoring the numerator gives x-intercepts $(-1,0)$ and $(0,0)$ ; the y-intercept is also $(0,0)$ . Equal leading coefficients give horizontal asymptote $y=1$ . Solving $u(x)=1$ gives $x=1$ , an allowed input, hence the crossing $(1,1)$ . Since $x^2+1>0$ , the remainder is negative for $x<1$ and positive for $x>1$ : this really is a crossing, not merely a touch. End behaviour does not forbid a finite crossing.

4. Build a formula from graph features

Construct one rational function with vertical asymptote $x=-2$ , horizontal asymptote $y=2$ , x-intercept $(3,0)$ , and a hole at $(1,-4/3)$ . State its domain and y-intercept. Your task is to give a valid formula, not to prove that the features determine a unique function.

Hint 4

Start with $2(x-3)/(x+2)$ , then introduce a common factor that excludes 1 without changing allowed outputs.

Explained solution 4

One answer is $v(x)=\frac{2(x-3)(x-1)}{(x+2)(x-1)},\qquad x\ne-2,1.$ The remaining denominator gives the vertical asymptote $x=-2$ ; the leading-coefficient ratio gives $y=2$ . The allowed zero is 3. At the cancelled input 1, the simplified expression gives $2(1-3)/(1+2)=-4/3$ , the required missing point. Zero is allowed, and $v(0)=-3$ , so the y-intercept is $(0,-3)$ . Writing only the simplified formula on its natural domain would wrongly fill the hole.

Choose your retry: if you restored an excluded input, repeat problem 2; if you confused a hole with unbounded behaviour, compare example 1 and problem 1; if you thought an asymptote was uncrossable, repeat problem 3. Bring your factored expression, domain and labelled sketch to a tutoring session. Tutoring is currently offered in English.

MATH 201 topics to review

Function notation and compositionGraph transformationsPolynomial and rational functionsExponential functionsLogarithmic functionsTrigonometric functions

Common difficulties

Tracking the domain through a composition
Distinguishing an inverse function from a reciprocal

MATH 201 review: early functions and later trigonometry

For weeks 1–6 preparation, start with the domain, inverse-function and logarithm checks, then the rational-graph workshop. Keep the fourth check for later trigonometry. Show your steps before opening the solutions.

The Fall 2026 outline places the midterm on weeks 1–6. Trigonometry begins in week 7, so the fourth check is later-course practice, not part of that early scope. Confirm your section’s current instructions. This selection is not a complete mock exam.

Before you start: Allow 25 minutes for a first attempt, then 15 minutes to check and retry any missed steps.

1. Combine two restrictions

Find the real domain of $r(x)=\sqrt{x+2}/(x-1)$ . Decide separately whether $x=-2$ and $x=1$ are allowed.

Show the explained solution

The square root requires $x+2\ge0$ , hence $x\ge-2$ . The denominator requires $x\ne1$ . Both conditions must hold at once.
The domain is $[-2,1)\cup(1,\infty).$ The input $-2$ is allowed because the numerator is zero and the denominator is $-3$ . The input $1$ is excluded because division by zero is undefined.
If you excluded the square-root endpoint, distinguish a nonnegative square-root argument from the strictly positive argument required by a logarithm.

Review algebra and domain restrictions

2. Read a transformation and invert one branch

Let $h(x)=2(x-3)^2-5$ for $x\ge3$ . Describe the transformations of $y=x^2$ , then find the inverse and its domain.

Show the explained solution

The graph moves three units right, stretches vertically by a factor of two, then moves five units down. The specified domain keeps the right-hand branch, where the function is one-to-one.
The range of $h$ is $[-5,\infty)$ . From $t=2(x-3)^2-5$ , use $x-3\ge0$ to take the nonnegative square root:
$h^{-1}(t)=3+\sqrt{\frac{t+5}{2}},\qquad t\ge-5.$ Its range is $[3,\infty)$ . The inverse’s domain comes from the original range.
At $t=3$ , the inverse returns $x=5$ , and $h(5)=3$ . Keeping both square-root signs would fail to define the inverse of the selected branch.

Review the worked composition and inverse example

3. Reject an algebraic root that violates a logarithm

Solve $\ln(x-1)+\ln(x+1)=\ln8$ over the reals. Here $\ln$ means natural logarithm.

Show the explained solution

Both arguments must be positive: $x-1>0$ and $x+1>0$ , giving the combined restriction $x>1$ .
On that domain, the logarithm product rule gives $\ln[(x-1)(x+1)]=\ln8$ . Since the logarithm is one-to-one, $x^2-1=8$ .
The algebraic candidates are $x=3$ and $x=-3$ . Only $x=3$ satisfies $x>1$ . Check: $\ln2+\ln4=\ln8$ .
Write the domain before combining logarithms. A correct quadratic calculation can still produce an inadmissible candidate.

Review logarithms and inverse operations

4. Later-course trigonometry: keep the solutions lost by division

Solve $2\sin^2\theta-\sin\theta=0$ for $0\le\theta<2\pi$ , using radians.

Show the explained solution

Factor rather than divide by $\sin\theta$ : $\sin\theta(2\sin\theta-1)=0.$
The first factor gives $\sin\theta=0$ , hence $\theta=0,\pi$ in the specified interval. The endpoint $2\pi$ is excluded.
The second factor gives $\sin\theta=1/2$ . Its reference angle is $\pi/6$ , and sine is positive in quadrants I and II, giving $\pi/6$ and $5\pi/6$ .
The complete set is $\{0,\pi/6,5\pi/6,\pi\}$ . Dividing by $\sin\theta$ at the start would discard the solutions $0$ and $\pi$ .

Review all trigonometric solutions on an interval

Use your answers to plan your revision

Domain error in question 1 or 3
Write one restriction per operation, intersect the restrictions, then check a boundary and an excluded value. Redo that question before starting another full set.
Inverse error in question 2
List domain and range side by side. Swap those sets for the inverse, then verify one composition and one numeric input.
Missing trigonometric solutions in question 4
Factor first, solve every factor, then filter by the interval. Practise one assigned equation with two different trigonometric branches.

For a lesson, you can bring any of these questions even if you have not started it. If you have attempted work or a current outline, include it. Your tutor can help you choose a starting point and work through functions, algebraic restrictions or trigonometry with you.

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Original Nablio practice, not an official examination.

Course topic sources

Preparing for this course

Review factoring, equation solving and coordinate graphs. Function composition, inverses and domains build on those tools.

Registration requirements

No formal course prerequisite is listed in the current calendar; MATH 200 is useful preparation, not a stated mandatory prerequisite.

Check the official course catalogue

Try a course problem

For a real input $x$ , simplify $\frac{x^2-16}{x-4}$ and state which input value must remain excluded from its domain.

Show the reasoning

The denominator is zero at $x=4$ , so first require $x\ne4$ . Factor the difference of squares: $x^2-16=(x-4)(x+4)$ . For the allowed inputs only, divide numerator and denominator by the nonzero factor $x-4$ . The result is $x+4$ , still with $x\ne4$ .

Composition order changes the function and its domain

The same two formulas can produce different composite functions. This example follows the input through each operation, then uses the range to determine an inverse function’s domain.

1. Apply the inner function first

Let $x$ be a real input. We use the nonnegative real square root. The notation $f\circ g$ means apply $g$ first, then apply $f$ to that output; name this composite function $h$ . Let $f(x)=\sqrt{x-1}$ and $g(x)=3x+2$ . Then $h(x)=(f\circ g)(x)=\sqrt{3x+1}.$ The square root requires $3x+1\ge0$ , so $h$ has domain (allowed inputs) $[-\tfrac13,\infty)$ and range (possible outputs) $[0,\infty)$ . The input to $f$ is $g(x)$ , not the original $x$ .

2. Reverse the order and check again

The reverse composition is $k(x)=(g\circ f)(x)=3\sqrt{x-1}+2.$ Here the square root happens first, so the domain is $[1,\infty)$ . Its range is $[2,\infty)$ . Thus $f\circ g$ and $g\circ f$ differ in both formula and allowable inputs.

3. Find the inverse with its input restriction

An inverse reverses input and output. The function $h$ is strictly increasing on its domain, so each value in its range comes from exactly one input. Write $y$ for the output of $h$ ; after solving for the original input, use $t$ to name an input to the inverse. For $h$ , start from $y=\sqrt{3x+1}$ , where $y\ge0$ . Solving gives $x=(y^2-1)/3$ . Therefore $h^{-1}(t)=\frac{t^2-1}{3},\qquad t\ge0.$ The inverse’s domain is the original function’s range; its range is $[-\tfrac13,\infty)$ .

4. Check what squaring can hide

Composing back gives $h(h^{-1}(t))=\sqrt{t^2}=|t|.$ This equals $t$ on the required domain $t\ge0$ . If you incorrectly allow $t=-2$ , the expression returns $2$ , not $-2$ . The restriction is essential, even though the inverse formula’s polynomial expression can be evaluated at negative numbers.

Try it yourself

Find $k^{-1}$ for $k(x)=3\sqrt{x-1}+2$ . State its domain and range, and explain the restriction.

Show the explained solution

From $t=3\sqrt{x-1}+2$ , obtain $\begin{gathered}k^{-1}(t)=1+\left(\frac{t-2}{3}\right)^2,\\t\ge2.\end{gathered}$ Its range is $[1,\infty)$ . The input restriction $t\ge2$ comes from the range of $k$ ; it also makes $(t-2)/3$ nonnegative before squaring.

For a new composition or inverse, write the formula, domain and range together. Test one boundary input and one excluded input to see whether the restrictions make sense.

How tutoring can help

You can bring a function or equation even if you do not know where to start. Your tutor can help you identify the allowed inputs, sketch the graph and work through a transformation or composition. For logarithms and inverse functions, you can work through the inverse operations together before trying another equation.

Exam preparation for MATH 201

For each function family, practise formula, domain, sketch and transformation. Mix logarithmic and trigonometric equations so you must choose the method.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Working with triangle lengths and angles?

Start by checking whether a right angle is given. For a right triangle, review choosing sine, cosine or tangent. For a triangle without a right angle, use the sine and cosine laws: match opposite sides and angles, choose a method from the given data, and check whether two different triangles are possible.

These are later-course topics in the Fall 2026 outlines. Use your section’s announced exam scope to decide when to practise them. After the lesson, try a new question without its solution and explain why your chosen method applies.

Independent practice

Free practice and study guides for MATH 201

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MATH 201 · official course reference MATH 201 · official course outline

Concordia MATH 201: Elementary Functions tutoring

Course overview

Start here

Rational graphs without calculus: keep the missing inputs

A five-step graph plan

Worked example 1: one hole and one vertical asymptote

Worked example 2: an oblique asymptote from division

Four original problems: predict, sketch, then check

1. A repeated denominator factor

2. Cancellation does not restore the y-intercept

3. Cross a horizontal asymptote

4. Build a formula from graph features

MATH 201 topics to review

Common difficulties

MATH 201 review: early functions and later trigonometry

1. Combine two restrictions

2. Read a transformation and invert one branch

3. Reject an algebraic root that violates a logarithm

4. Later-course trigonometry: keep the solutions lost by division

Preparing for this course

Registration requirements

Try a course problem

Composition order changes the function and its domain

1. Apply the inner function first

2. Reverse the order and check again

3. Find the inverse with its input restriction

4. Check what squaring can hide

Try it yourself

How tutoring can help

Exam preparation for MATH 201

Working with triangle lengths and angles?

Free practice and study guides for MATH 201

Cancelling Factors: Why the Missing Input Stays Missing

Before Calculus: 10 Algebra Skills You Should Know

Concordia MATH 203 Readiness Checklist

Exponential and Logarithmic Functions

Functions: Transformations, Composition and Inverses

Unit Circle and Trigonometric Graphs

Need help with MATH 201?

Concordia MATH 201: Elementary Functions tutoring

Course overview

Start here

Rational graphs without calculus: keep the missing inputs

A five-step graph plan

Worked example 1: one hole and one vertical asymptote

Worked example 2: an oblique asymptote from division

Four original problems: predict, sketch, then check

1. A repeated denominator factor

2. Cancellation does not restore the y-intercept

3. Cross a horizontal asymptote

4. Build a formula from graph features

MATH 201 topics to review

Common difficulties

MATH 201 review: early functions and later trigonometry

1. Combine two restrictions

2. Read a transformation and invert one branch

3. Reject an algebraic root that violates a logarithm

4. Later-course trigonometry: keep the solutions lost by division

Preparing for this course

Registration requirements

Try a course problem

Composition order changes the function and its domain

1. Apply the inner function first

2. Reverse the order and check again

3. Find the inverse with its input restriction

4. Check what squaring can hide

Try it yourself

How tutoring can help

Exam preparation for MATH 201

Working with triangle lengths and angles?

Related courses

Free practice and study guides for MATH 201

Cancelling Factors: Why the Missing Input Stays Missing

Before Calculus: 10 Algebra Skills You Should Know

Concordia MATH 203 Readiness Checklist

Exponential and Logarithmic Functions

Functions: Transformations, Composition and Inverses

Unit Circle and Trigonometric Graphs

Need help with MATH 201?