Source: MTH 132 Chapter 3 Lecture Notes, Michigan State University
Difficulty: Intermediate to Advanced | Prerequisites: All of Chapter 3 Sections 1 through 3 (critical numbers, EVT, MVT, increasing/decreasing, concavity, inflection points), polynomial long division, vertical and horizontal asymptotes.
Curve sketching pulls together every tool from Chapter 3 into one task: draw a function's graph by hand using calculus rather than point-plotting. You will also encounter slant asymptotes, which appear when the degree of the numerator is exactly one more than the denominator. Optimisation problems take everything further: you translate a real-world scenario into a function and then find its absolute maximum or minimum. These are the payoff problems for the entire chapter.
Curve sketching means systematically finding a function's domain, intercepts, symmetry, asymptotes, intervals of increase/decrease, local extrema, concavity, and inflection points, then combining all of that into a sketch. Slant asymptotes arise from polynomial long division. Optimisation problems model real situations as functions and use calculus to find the best outcome.
Slant (oblique) asymptote
A line y = mx + b that the graph of f(x) approaches as x tends to positive or negative infinity. Occurs when the degree of the numerator is exactly one more than the degree of the denominator. Found by polynomial long division.
Polynomial long division
The same division algorithm you learned for numbers, applied to polynomials. Used to rewrite a rational function as a linear quotient plus a remainder fraction.
Optimisation problem
A word problem where you must find the maximum or minimum value of some quantity (area, volume, cost, distance) subject to a constraint. You set up a function of one variable and apply the closed interval method or the First Derivative Test.
A rational function f(x) = p(x)/q(x) has a slant asymptote when the degree of p is exactly one more than the degree of q. To find it:
Perform polynomial long division of p(x) by q(x).
The quotient (ignoring the remainder) is the slant asymptote y = mx + b.
You can verify: lim as x approaches infinity of [f(x) - (mx + b)] = 0.
Example: f(x) = (2x^2 - x)/(x + 1)
Divide 2x^2 - x by x + 1. Quotient: 2x - 3, remainder: 3.
f(x) = 2x - 3 + 3/(x + 1). As x approaches infinity, 3/(x + 1) approaches 0.
Slant asymptote: y = 2x - 3.
Example: y = (x^2 + 3x + 2)/(x - 2)
Divide x^2 + 3x + 2 by x - 2. Quotient: x + 5, remainder: 12.
Slant asymptote: y = x + 5.
For a function f(x), work through these steps:
Domain. Where is f defined?
Intercepts. Set x = 0 for the y-intercept; set f(x) = 0 for x-intercepts.
Asymptotes. Vertical (where denominator = 0), horizontal (limits as x approaches infinity), slant (polynomial long division if degree of numerator is one more).
First derivative analysis. Find f'(x), its critical numbers, and build a sign chart. Determine intervals of increase/decrease and classify critical points using the First Derivative Test.
Second derivative analysis. Find f''(x), determine where it is zero or undefined, and build a sign chart. Determine intervals of concavity and identify inflection points.
Sketch. Plot the key points (intercepts, critical points, inflection points), draw the asymptotes, and connect with a curve that respects the increasing/decreasing and concavity information.
Worked example: f(x) = x(x - 4)^3
Domain: all real numbers.
y-intercept: f(0) = 0. x-intercepts: x = 0 and x = 4.
No vertical, horizontal, or slant asymptotes (polynomial).
f'(x) = 4(x - 4)^2(x - 1). Critical numbers: x = 1, x = 4.
f' < 0 on (-infinity, 1): decreasing. f' > 0 on (1, 4) and (4, infinity): increasing.
Local min at x = 1. Neither at x = 4 (f' does not change sign).
f''(x) = 12(x - 4)(x - 2). f'' = 0 at x = 2 and x = 4.
Concave up on (-infinity, 2) and (4, infinity). Concave down on (2, 4).
Inflection points at x = 2 and x = 4.
Optimisation problems are what related rates are to implicit differentiation: the applied payoff. The general approach:
Draw a picture and introduce variables.
Write the quantity to optimise as a function (area, volume, cost, etc.).
Use the constraint to eliminate one variable, getting a function of a single variable.
Find the domain of the resulting function (what values make physical sense).
Apply calculus (closed interval method, or First Derivative Test) to find the absolute max or min.
Answer the question in context.
Example: Find the maximum area of a rectangle inscribed in an equilateral triangle with side length 6, where one side of the rectangle lies along the base.
Let w be the width of the rectangle. The height of the equilateral triangle is 3sqrt(3). Using the slope of the triangle's side, the height of the rectangle is h = -sqrt(3)/2 * w + 3sqrt(3).
Area = w * h = w(-sqrt(3)/2 * w + 3sqrt(3)) = -sqrt(3)/2 * w^2 + 3sqrt(3) * w.
Domain: w is in [0, 6].
A'(w) = -sqrt(3) * w + 3*sqrt(3). Set A' = 0: w = 3.
A(0) = 0, A(3) = 9*sqrt(3)/2, A(6) = 0.
Maximum area = 9*sqrt(3)/2 square units, achieved when w = 3.
Students often forget to check for slant asymptotes. If the numerator's degree is exactly one more than the denominator's, you have one. If the degrees are equal, you have a horizontal asymptote instead.
In optimisation, a common error is optimising a function of two variables without using the constraint to reduce to one variable.
Students sometimes skip checking the endpoints in optimisation problems. The absolute max or min can occur at the boundary of the feasible domain.
In curve sketching, students forget that the absence of a vertical asymptote does not mean the function has no asymptotes. Check horizontal and slant as well.
Curve sketching problems are long, multi-part exam questions that test everything at once: domain, intercepts, asymptotes, first derivative analysis, second derivative analysis, and putting it all together.
Polynomial long division for slant asymptotes is tested separately and within curve sketching.
Optimisation word problems are a staple of Calculus I exams. Know the general procedure cold: draw, write, constrain, differentiate, solve, check endpoints.
True or False: A rational function always has a slant asymptote.
Answer: False. A slant asymptote occurs only when the numerator's degree is exactly one more than the denominator's.
Fill in the blank: To find a slant asymptote, perform ____ of the numerator by the denominator.
Answer: polynomial long division.
True or False: In an optimisation problem, if the domain is a closed interval, the absolute max or min must occur at a critical number or an endpoint.
Answer: True.
Q: Find the slant asymptote of y = (2x^2 - x)/(x + 1).
A: Polynomial long division gives 2x - 3 with remainder 3. Slant asymptote: y = 2x - 3.
Q: For f(x) = x(x - 4)^3, find the domain, x-intercepts, intervals of increase/decrease, local extrema, intervals of concavity, and inflection points.
A: Domain: all reals. x-intercepts: x = 0, x = 4. f'(x) = 4(x-4)^2(x-1). Decreasing on (-infinity, 1), increasing on (1, 4) and (4, infinity). Local min at x = 1, neither at x = 4. f''(x) = 12(x-4)(x-2). Concave up on (-infinity, 2) and (4, infinity), concave down on (2, 4). Inflection points at x = 2 and x = 4.
Q: A rectangle is inscribed in an equilateral triangle with side 6, with one side on the base. Find the maximum area.
A: Let w = width. Height h = -sqrt(3)/2 * w + 3sqrt(3). Area A = w * h. A'(w) = -sqrt(3)w + 3sqrt(3) = 0 gives w = 3. A(3) = 9sqrt(3)/2. This is the maximum (endpoints give area = 0).
Q: Find the slant asymptote of y = (x^2 + 3x + 2)/(x - 2).
A: Divide x^2 + 3x + 2 by x - 2. x - 2 goes into x^2 + 3x + 2: first term x (since x times (x-2) = x^2 - 2x), remainder 5x + 2. Then 5 (since 5 times (x-2) = 5x - 10), remainder 12. So y = x + 5 + 12/(x - 2). Slant asymptote: y = x + 5.
Q: Use polynomial long division to write 219/12 as a mixed number.
A: 12 goes into 219 eighteen times (12 x 18 = 216), remainder 3. So 219/12 = 18 + 3/12 = 18 + 1/4. This is the same logic used for polynomials: divide, multiply, subtract, bring down.
Q: Outline the curve sketching steps for f(x) = (2x^2 - x)/(x + 1).
A: Domain: all x except x = -1. y-intercept: f(0) = 0. x-intercepts: 2x^2 - x = x(2x - 1) = 0, so x = 0 and x = 1/2. Vertical asymptote at x = -1. Slant asymptote: long division gives y = 2x - 3 (remainder 3/(x+1)). Then compute f' and f'' to find intervals of increase/decrease, concavity, local extrema, and inflection points. Sketch by connecting everything.
Q: A farmer has 200 metres of fencing and wants to enclose a rectangular area next to a river (no fence needed along the river). What dimensions maximise the area?
A: Let x = length parallel to river, y = width perpendicular. Constraint: x + 2y = 200, so x = 200 - 2y. Area A = xy = (200 - 2y)y = 200y - 2y^2. Domain: y in [0, 100]. A'(y) = 200 - 4y = 0 gives y = 50. A(0) = 0, A(50) = 5000, A(100) = 0. Maximum area = 5000 square metres, with y = 50 m and x = 100 m.
Curve sketching integrates every concept from Chapter 3 (critical numbers, EVT, increasing/decreasing, concavity, asymptotes) into a single problem type.
Optimisation connects to related rates (Chapter 2/3) in the sense that both require translating word problems into calculus. Related rates involve time derivatives; optimisation involves finding extrema.
Slant asymptotes reappear in rational function graphing in precalculus and in more advanced calculus topics involving limits at infinity.
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