Difficulty: Intermediate | Prerequisites: Critical points and the second derivative test (previous notes), limits at infinity
Graph sketching brings together everything from Chapter 3: critical points, the second derivative test, concavity, and now asymptotes. The goal is to draw an accurate picture of a function's behaviour without plotting dozens of points. This is the payoff section, where all the individual tools combine into one systematic procedure. If you can find critical points and compute limits, you are ready.
Asymptotes are invisible boundary lines that a graph approaches but never (or rarely) crosses. Vertical asymptotes come from values that make the denominator zero; horizontal asymptotes describe where the function settles as x goes to positive or negative infinity. Graph sketching is the process of combining asymptotes, intercepts, critical points, and concavity into one coherent picture of the function.
Vertical asymptote (VA)
A vertical line x = a that the graph approaches but never touches, where f(x) tends toward positive or negative infinity. Found by setting the denominator of a rational function equal to zero (after cancelling any common factors with the numerator).
Think of it as: a wall the graph zooms toward but cannot pass through.
Horizontal asymptote (HA)
A horizontal line y = L that the graph approaches as x goes to positive infinity, negative infinity, or both. Found by evaluating the limits of f(x) as x approaches ±∞.
In simple terms, this means: the long-run value the function settles toward when x gets very large or very small.
Increasing function
A function is increasing on an interval if f'(x) > 0 throughout that interval. The graph rises from left to right.
Decreasing function
A function is decreasing on an interval if f'(x) < 0 throughout that interval. The graph falls from left to right.
Inflection point
A point where the concavity of the graph changes (from concave up to concave down, or vice versa). Found where f''(x) = 0 or is undefined, provided the concavity actually switches.
Think of it as: the spot where the curve stops bending one way and starts bending the other.
Sign chart (sign diagram)
A number-line diagram used to track where a derivative is positive or negative across different intervals. You mark the critical points (or zeros of f''), then test one value in each interval.
In simple terms, this means: a quick visual that tells you where the function goes up, goes down, bends up, or bends down.
x-intercept
The point(s) where the graph crosses the x-axis. Found by setting f(x) = 0 and solving for x.
y-intercept
The point where the graph crosses the y-axis. Found by evaluating f(0).
Factor both the numerator and denominator of the rational function.
Cancel any common factors (those produce holes, not asymptotes).
Set the remaining denominator equal to zero and solve. Each solution is a vertical asymptote.
Example: f(x) = (x² − 9) / (x² + 2x)
Factor: (x + 3)(x − 3) / (x(x + 2))
No common factors to cancel.
Denominator = 0: x = 0 and x = −2
Vertical asymptotes at x = 0 and x = −2
Compare the degrees of numerator and denominator:
If degree of numerator < degree of denominator: HA is y = 0
If degree of numerator = degree of denominator: HA is y = (leading coefficient of numerator) / (leading coefficient of denominator)
If degree of numerator > degree of denominator: no horizontal asymptote
Alternatively, evaluate lim x→∞ f(x) and lim x→−∞ f(x) directly.
Example: f(x) = x² / (x² + 1)
Same degree (both 2). Leading coefficients: 1/1 = 1
HA: y = 1
Confirm: lim x→∞ [x² / (x² + 1)] = 1, and lim x→−∞ [x² / (x² + 1)] = 1
For any function, work through these steps in order:
Find the domain
Find vertical asymptotes (rational functions)
Find horizontal asymptotes (rational functions)
Find x-intercepts: set f(x) = 0
Find y-intercept: evaluate f(0)
Find f'(x), set it to zero to get critical points; build a sign chart for increasing/decreasing intervals
Find f''(x), set it to zero to get possible inflection points; build a sign chart for concavity
Plot the key points and sketch the curve, respecting all of the above
Vertical asymptote: (x + 1)² = 0, so x = −1
Horizontal asymptote: Degree of numerator (1) < degree of denominator (2), so HA is y = 0
First derivative (critical points):
Using the quotient rule:
f'(x) = [(−1)(x + 1)² − (−x)(2)(x + 1)] / (x + 1)⁴
Simplify the numerator: −(x + 1) + 2x = x − 1
So f'(x) = (x − 1) / (x + 1)³
Set f'(x) = 0: x − 1 = 0, so x = 1 is the critical point.
Sign chart for f'(x):
For x < −1: test x = −2, f'(−2) = (−3)/(−1)³ = 3 > 0 (increasing)
For −1 < x < 1: test x = 0, f'(0) = (−1)/(1)³ = −1 < 0 (decreasing)
For x > 1: test x = 2, f'(2) = (1)/(3)³ = 1/27 > 0 (increasing)
Since f' changes from negative to positive at x = 1, there is a relative minimum at x = 1.
f(1) = −1/(1 + 1)² = −1/4, so the relative minimum is at (1, −1/4).
Second derivative (concavity):
f''(x) = 2(x − 2) / (x + 1)⁴
Set f''(x) = 0: x = 2
Sign chart for f''(x):
For x < −1: concave down
For −1 < x < 2: concave down
For x > 2: concave up
Inflection point at x = 2 (concavity changes from down to up).
Intercepts:
x-intercept: −x / (x + 1)² = 0, so x = 0. The graph passes through (0, 0).
y-intercept: f(0) = 0, confirming (0, 0).
Asymptotes: No denominator, so no vertical asymptotes. As x → ±∞, x³ → ±∞, so no horizontal asymptotes.
First derivative: f'(x) = 3x². Setting f'(x) = 0 gives x = 0. Since 3x² ≥ 0 for all x, f is never decreasing (it is increasing everywhere except at x = 0 where the slope is momentarily zero).
Second derivative: f''(x) = 6x. Setting f''(x) = 0 gives x = 0.
For x < 0: f''(x) < 0 (concave down)
For x > 0: f''(x) > 0 (concave up)
Inflection point at (0, 0), where the concavity switches.
Intercepts: x-intercept at x = 0, y-intercept at (0, 0). The origin is both the intercept and the inflection point.
\text{VA: set denominator} = 0 \text{ (after cancelling common factors)}\text{HA: } \lim_{x \to \pm\infty} f(x) = L \quad \Rightarrow \quad y = L\text{Quotient rule: } \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}f'(x) > 0 \Rightarrow \text{increasing} \qquad f'(x) < 0 \Rightarrow \text{decreasing}f''(x) > 0 \Rightarrow \text{concave up} \qquad f''(x) < 0 \Rightarrow \text{concave down}Graph sketching is how economists visualise cost, revenue, and profit curves to identify break-even points and optimal output levels. Asymptotes appear in models of diminishing returns (e.g. average cost per unit approaching a floor as production scales up, but never quite reaching it) and in pharmacokinetics, where drug concentration in the blood approaches zero over time but the mathematical model never hits exactly zero.
Students often think a function can never cross a horizontal asymptote. A function can cross its HA in finite regions of the graph; the HA only describes end behaviour as x approaches infinity.
Students sometimes forget to cancel common factors before identifying vertical asymptotes. If numerator and denominator share a factor, that x-value is a hole (removable discontinuity), not an asymptote.
Students frequently skip the sign chart and try to guess the shape of the graph from the critical points alone. Without checking the sign of f' and f'' in each interval, you cannot know which direction the curve moves between key points.
Students sometimes assume that f''(x) = 0 always produces an inflection point. It does not. The concavity must actually change at that point. Check the sign of f'' on both sides.
⚠️ "Sketch the graph of f(x)" is a multi-part question that tests nearly everything in the chapter at once. Expect it on every exam covering Chapter 3.
⚠️ Exams often give a rational function with a repeated factor in the denominator (like (x + 1)²) to test whether you handle the sign of f(x) near the asymptote correctly. The graph approaches the same direction from both sides when the factor is squared.
⚠️ You will likely be asked to find both VAs and HAs separately, then use them again in a full graph sketch. Practise the limit calculations until they are automatic.
⚠️ Sign charts are frequently required as shown work. Even if the question says only "sketch," many professors expect the sign chart for full marks.
True or false: If the denominator of a rational function equals zero at x = 3, then x = 3 is always a vertical asymptote. (False: if the numerator also equals zero at x = 3 and the factor cancels, it is a hole.)
Fill in the blank: When the degree of the numerator equals the degree of the denominator, the horizontal asymptote is y = ________. (ratio of leading coefficients)
True or false: A function can have at most two horizontal asymptotes. (True: one as x → ∞ and one as x → −∞.)
Fill in the blank: f'(x) > 0 on an interval means f is ________ on that interval. (increasing)
True or false: f(x) = x³ has a relative minimum at x = 0. (False: x = 0 is an inflection point, not an extremum.)
Q: Find all vertical and horizontal asymptotes of f(x) = (x² − 9) / (x² + 2x).
A: Factor: (x + 3)(x − 3) / [x(x + 2)]. No common factors. VA: x = 0, x = −2. For the HA: degrees are equal, leading coefficients are both 1, so HA: y = 1.
Q: Determine the horizontal asymptote of f(x) = 3x / (x² + 5).
A: Degree of numerator (1) < degree of denominator (2), so HA is y = 0.
Q: For f(x) = −x / (x + 1)², find the intervals where f is increasing and decreasing.
A: f'(x) = (x − 1) / (x + 1)³. Critical point at x = 1. VA at x = −1. Increasing on (−∞, −1) and (1, ∞). Decreasing on (−1, 1).
Q: Sketch the key features of f(x) = x³. State the intercepts, any asymptotes, intervals of increase/decrease, and concavity.
A: No asymptotes. x-intercept and y-intercept both at (0, 0). f'(x) = 3x² ≥ 0, so f is increasing everywhere (flat at x = 0). f''(x) = 6x: concave down on (−∞, 0), concave up on (0, ∞). Inflection point at (0, 0).
Q: A function has a vertical asymptote at x = 2. Its denominator contains the factor (x − 2)². Does the graph go to +∞ on both sides of x = 2, or opposite directions?
A: Because the factor is squared (even power), the denominator is positive on both sides of x = 2. The graph goes the same direction (both to +∞ or both to −∞) depending on the sign of the numerator at x = 2.
This connects to the previous topic on critical points and the second derivative test, because graph sketching is where you use those tools together with asymptotes and intercepts to build the full picture. It also connects to limits (Chapter 1 or 2 in most courses), since horizontal asymptotes are defined by limits at infinity. In later chapters, these sketching skills feed directly into applied optimisation problems, where drawing the curve helps you confirm that your algebraic answer makes sense.
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