Math Analysis 1

Solved curve-sketching exercises, with graphs

Twelve complete function studies with graphs: polynomials, rational functions, exponentials, logarithms, roots, absolute values, cusps, oblique asymptotes and number of solutions of an equation with a parameter.

ielettronica 25 October 2026 13 min read 0 comments

Curve-sketching exercises (studying the graph of a function) of the kind set in Calculus 1 (Mathematical Analysis 1) courses, in three levels: basic, with polynomials, rational functions and the exponential; intermediate, with logarithms, roots and absolute values; exam, with oblique asymptotes, cusps, split domains and discussion of the number of solutions of an equation, as in written exams. For each function the result summarises the key points; the complete solution, step by step and with the graph, opens with a click. The proofs of the criteria used are in the in-depth study linked at the top.

In the graphs the red dots are maxima and minima, the green ones inflection points, the black ones intersections with the axes; asymptotes are dashed in red.

Basic level

Exercise 1. A cubic polynomial

Study $f(x) = x^3 - 3x$ and sketch its graph.

Result: odd; local maximum $(-1, 2)$ , local minimum $(1, -2)$ , inflection point $(0, 0)$ ; no asymptotes.

Solution

1. Domain and symmetry. $\mathbb R$ . $f(-x) = -x^3 + 3x = -f(x)$ : the function is odd, the graph is symmetric about the origin.

2. Zeros and sign. $x(x^2 - 3) = 0$ for $x = 0$ and $x = \pm\sqrt 3 \approx \pm 1.73$ . $f \gt 0$ for $-\sqrt 3 \lt x \lt 0$ and for $x \gt \sqrt 3$ .

3. Limits and asymptotes. $f \to \pm\infty$ as $x \to \pm\infty$ . No asymptotes: $f(x)/x = x^2 - 3 \to +\infty$ .

4. First derivative. $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ : positive for $|x| \gt 1$ , negative for $|x| \lt 1$ . Increasing on $(-\infty, -1)$ and $(1, +\infty)$ , decreasing on $(-1, 1)$ . Local maximum $f(-1) = 2$ , local minimum $f(1) = -2$ .

5. Second derivative. $f''(x) = 6x$ : concave for $x \lt 0$ , convex for $x \gt 0$ . Inflection point at the origin, with tangent of slope $f'(0) = -3$ .

6. Graph. The maximum and minimum are only local, because the function is unbounded.

Graph of x cubed minus 3 x — Local maximum at minus 1, local minimum at 1, inflection point at the origin; the graph crosses the x axis at minus root 3, 0 and root 3.

Exercise 2. A rational function with a horizontal asymptote

Study $f(x) = \dfrac{x}{x^2 + 1}$ .

Result: odd; asymptote $y = 0$ ; absolute minimum $\left(-1, -\frac{1}{2}\right)$ , absolute maximum $\left(1, \frac{1}{2}\right)$ ; inflection points at $x = 0$ and $x = \pm\sqrt 3$ .

Solution

1. Domain and symmetry. The denominator never vanishes: domain $\mathbb R$ . Odd.

2. Zeros and sign. Zero at $x = 0$ ; sign of $x$ .

3. Limits and asymptotes. Numerator degree lower than denominator degree: $f \to 0$ as $x \to \pm\infty$ , horizontal asymptote $y = 0$ .

4. First derivative. $f'(x) = \dfrac{(x^2 + 1) - x\cdot 2x}{(x^2 + 1)^2} = \dfrac{1 - x^2}{(x^2 + 1)^2}$ . Positive for $|x| \lt 1$ . Decreasing on $(-\infty, -1)$ , increasing on $(-1, 1)$ , decreasing on $(1, +\infty)$ . Minimum $f(-1) = -\frac{1}{2}$ , maximum $f(1) = \frac{1}{2}$ . They are absolute: the function tends to 0 at the ends and never exceeds these values.

5. Second derivative. $f''(x) = \dfrac{2x(x^2 - 3)}{(x^2 + 1)^3}$ . The sign is that of $x(x^2 - 3)$ : negative for $x \lt -\sqrt 3$ , positive between $-\sqrt 3$ and 0, negative between 0 and $\sqrt 3$ , positive beyond. Three inflection points: $(0, 0)$ and $\left(\pm\sqrt 3, \pm\frac{\sqrt 3}{4}\right)$ .

6. Graph. The graph crosses the asymptote at the origin: a horizontal asymptote only describes the behaviour at infinity.

Graph of x over x squared plus 1 — Minimum at minus 1 and maximum at 1, inflection points at minus root 3, 0 and root 3; the graph approaches the x axis on both sides.

Exercise 3. The Gaussian

Study $f(x) = e^{-x^2}$ .

Result: even, always positive; asymptote $y = 0$ ; absolute maximum $(0, 1)$ ; inflection points at $x = \pm\frac{1}{\sqrt 2}$ .

Solution

1. Domain and symmetry. $\mathbb R$ ; even.

2. Sign. An exponential is always positive: no zeros.

3. Limits. $-x^2 \to -\infty$ , so $f \to 0$ : asymptote $y = 0$ on both sides.

4. First derivative. $f'(x) = -2x\,e^{-x^2}$ : increasing for $x \lt 0$ , decreasing for $x \gt 0$ . Absolute maximum $f(0) = 1$ .

5. Second derivative. $f''(x) = (-2 + 4x^2)e^{-x^2} = 2(2x^2 - 1)e^{-x^2}$ : negative for $|x| \lt \frac{1}{\sqrt 2}$ , positive outside. Inflection points at $x = \pm\frac{1}{\sqrt 2} \approx \pm 0.707$ , where $f = e^{-1/2} \approx 0.607$ .

6. Graph. It is the bell curve of the normal distribution (up to constants): the inflection points lie one “standard deviation” from the centre, here $\sigma = \frac{1}{\sqrt 2}$ .

Graph of e to the minus x squared — The bell has its maximum at 0 and the two inflection points at plus and minus 1 over root 2; it tends to zero on both sides.

Intermediate level

Exercise 4. Vertical and oblique asymptotes

Study $f(x) = \dfrac{x^2}{x - 1}$ .

Result: asymptotes $x = 1$ and $y = x + 1$ ; local maximum $(0, 0)$ , local minimum $(2, 4)$ ; concave for $x \lt 1$ , convex for $x \gt 1$ , no inflection point.

Solution

1. Domain. $x \ne 1$ . No symmetry (the domain is not symmetric).

2. Zeros and sign. Double zero at $x = 0$ . The numerator is $\ge 0$ : the sign is that of $x - 1$ , negative to the left of 1 and positive to the right.

3. Limits and asymptotes. $x \to 1^\pm$ : $f \to \pm\infty$ , vertical asymptote $x = 1$ . Division: $x^2 = (x - 1)(x + 1) + 1$ , so $f(x) = x + 1 + \dfrac{1}{x - 1}$ : oblique asymptote $y = x + 1$ , with the graph above it for $x \gt 1$ and below for $x \lt 1$ .

4. First derivative. $f'(x) = \dfrac{2x(x - 1) - x^2}{(x - 1)^2} = \dfrac{x(x - 2)}{(x - 1)^2}$ . Positive for $x \lt 0$ and for $x \gt 2$ . Increasing on $(-\infty, 0)$ , decreasing on $(0, 1)$ and on $(1, 2)$ , increasing on $(2, +\infty)$ . Local maximum $f(0) = 0$ , local minimum $f(2) = 4$ .

The maximum is less than the minimum: this is not a contradiction, because they lie on two branches separated by the vertical asymptote.

5. Second derivative. From the form $f = x + 1 + (x - 1)^{-1}$ : $f''(x) = \dfrac{2}{(x - 1)^3}$ . Concave for $x \lt 1$ , convex for $x \gt 1$ . No inflection point: the change of concavity happens at $x = 1$ , which does not belong to the domain.

Graph of x squared over x minus 1 — Two branches separated by the vertical asymptote x equal to 1: the left one has a maximum at the origin, the right one a minimum at 2, 4. Both approach the oblique asymptote y equal to x plus 1.

Exercise 5. Logarithm over x

Study $f(x) = \dfrac{\ln x}{x}$ and use the result to decide which number is larger between $e^\pi$ and $\pi^e$ .

Result: asymptotes $x = 0$ and $y = 0$ ; absolute maximum $\left(e, \frac{1}{e}\right)$ ; inflection point at $x = e^{3/2}$ ; $e^\pi \gt \pi^e$ .

Solution

1. Domain. $x \gt 0$ .

2. Zeros and sign. Zero at $x = 1$ ; positive for $x \gt 1$ .

3. Limits. $x \to 0^+$ : $\ln x \to -\infty$ and $1/x \to +\infty$ , the product $\to -\infty$ : vertical asymptote $x = 0$ . $x \to +\infty$ : $\ln x/x \to 0$ (hierarchy of infinities), horizontal asymptote $y = 0$ .

4. First derivative. $f'(x) = \dfrac{\frac{1}{x}\cdot x - \ln x}{x^2} = \dfrac{1 - \ln x}{x^2}$ : positive for $x \lt e$ . Absolute maximum $f(e) = \frac{1}{e} \approx 0.368$ .

5. Second derivative. $f''(x) = \dfrac{2\ln x - 3}{x^3}$ : inflection point at $x = e^{3/2} \approx 4.48$ , $f = \frac{3}{2e^{3/2}} \approx 0.335$ ; concave before, convex after.

6. Application. Since the maximum is at $e$ , for $\pi \ne e$ we have $\frac{\ln\pi}{\pi} \lt \frac{\ln e}{e} = \frac{1}{e}$ . Multiplying by $e\pi \gt 0$ : $e\ln\pi \lt \pi$ , i.e. $\ln(\pi^e) \lt \ln(e^\pi)$ , hence $\pi^e \lt e^\pi$ (about 22.46 against 23.14). Curve sketching also serves to prove inequalities.

Graph of log x over x — It starts from minus infinity near the y axis, crosses the x axis at 1, has its maximum at x equal to e, the inflection point at e to the three halves and then slowly decreases towards zero.

Exercise 6. A root with vertical tangents

Study $f(x) = x\sqrt{4 - x^2}$ .

Result: domain $[-2, 2]$ , odd; absolute maximum $\left(\sqrt 2, 2\right)$ , absolute minimum $\left(-\sqrt 2, -2\right)$ ; inflection point $(0, 0)$ ; vertical tangents at $x = \pm 2$ .

Solution

1. Domain and symmetry. $4 - x^2 \ge 0$ , i.e. $-2 \le x \le 2$ . Odd.

2. Zeros and sign. Zeros at 0 and $\pm 2$ ; sign of $x$ .

3. Limits and asymptotes. The domain is a closed interval and $f$ is continuous: no asymptotes, and by the extreme value theorem it has an absolute maximum and minimum.

4. First derivative. For $-2 \lt x \lt 2$ :

$f'(x) = \sqrt{4 - x^2} + x\cdot\frac{-x}{\sqrt{4 - x^2}} = \frac{4 - 2x^2}{\sqrt{4 - x^2}}$

It vanishes at $x = \pm\sqrt 2$ . Increasing on $(-\sqrt 2, \sqrt 2)$ , decreasing on $(-2, -\sqrt 2)$ and on $(\sqrt 2, 2)$ . Maximum $f(\sqrt 2) = \sqrt 2\cdot\sqrt 2 = 2$ , minimum $-2$ .

At the endpoints: $\lim_{x\to 2^-}f'(x) = \frac{-4}{0^+} = -\infty$ . By the limit-of-the-derivative theorem, at $x = 2$ the tangent is vertical; the same at $-2$ .

5. Second derivative. $f''(x) = \dfrac{2x(x^2 - 6)}{(4 - x^2)^{3/2}}$ . In the domain $x^2 - 6 \lt 0$ , so the sign is opposite to that of $x$ : convex for $x \lt 0$ , concave for $x \gt 0$ . Inflection point at the origin, with slope $f'(0) = 2$ .

Graph of x times the square root of 4 minus x squared — The graph is defined between minus 2 and 2, where it arrives with a vertical tangent; minimum at minus root 2, maximum at root 2, inflection point at the origin.

Exercise 7. An absolute value and a corner point

Study $f(x) = |x|\,e^{-x}$ .

Result: absolute minimum $(0, 0)$ at a corner point; local maximum $\left(1, \frac{1}{e}\right)$ ; inflection point $\left(2, \frac{2}{e^2}\right)$ ; asymptote $y = 0$ as $x \to +\infty$ .

Solution

1. Domain. $\mathbb R$ . Split the absolute value: $f(x) = x\,e^{-x}$ for $x \ge 0$ , $f(x) = -x\,e^{-x}$ for $x \lt 0$ .

2. Sign. $f \ge 0$ , zero only at 0: the origin is therefore the absolute minimum, whatever the derivative says.

3. Limits. $x \to +\infty$ : $x/e^x \to 0$ , asymptote $y = 0$ . $x \to -\infty$ : $-x\,e^{-x} \to +\infty$ , and $f(x)/x = -e^{-x} \to -\infty$ : no oblique asymptote.

4. First derivative. For $x \gt 0$ : $f' = (1 - x)e^{-x}$ . For $x \lt 0$ : $f' = (x - 1)e^{-x} \lt 0$ . So $f$ decreases on $(-\infty, 0)$ , increases on $(0, 1)$ , decreases on $(1, +\infty)$ : local maximum $f(1) = \frac{1}{e}$ .

At 0: $f'(0^+) = 1$ , $f'(0^-) = -1$ . Finite and different one-sided derivatives: corner point.

5. Second derivative. For $x \gt 0$ : $f'' = (x - 2)e^{-x}$ , inflection point at $x = 2$ , $f(2) = 2/e^2 \approx 0.271$ . For $x \lt 0$ : $f'' = (2 - x)e^{-x} \gt 0$ , convex.

Graph of the absolute value of x times e to the minus x — On the left the graph falls from plus infinity to the origin, where it forms a corner; on the right it rises to the maximum at 1, changes concavity at 2 and flattens onto the x axis.

Exam level

Exercise 8. An asymptote on one side only

Study $f(x) = x\,e^{1/x}$ .

Result: vertical asymptote $x = 0$ from the right only; oblique asymptote $y = x + 1$ ; local minimum $(1, e)$ ; as $x \to 0^-$ the graph reaches the origin with a horizontal tangent; no inflection point.

Solution

1. Domain. $x \ne 0$ .

2. Sign. Sign of $x$ ; no zeros.

3. Limits. $x \to 0^+$ : with $t = 1/x \to +\infty$ , $f = e^t/t \to +\infty$ : vertical asymptote from the right. $x \to 0^-$ : $1/x \to -\infty$ , $e^{1/x} \to 0$ , and $f \to 0$ : from the left no asymptote, the graph approaches the origin.

Oblique: $m = \lim e^{1/x} = 1$ ; $q = \lim x\left(e^{1/x} - 1\right) = 1$ (with $t = 1/x$ , $\frac{e^t - 1}{t} \to 1$ ). Asymptote $y = x + 1$ on both sides. With Taylor, $f(x) - (x + 1) = x\left(e^{1/x} - 1 - \frac{1}{x}\right) \sim x\cdot\frac{1}{2x^2} = \frac{1}{2x}$ : graph above the asymptote as $x \to +\infty$ , below as $x \to -\infty$ .

4. First derivative. $f'(x) = e^{1/x} + x\,e^{1/x}\left(-\frac{1}{x^2}\right) = e^{1/x}\,\dfrac{x - 1}{x}$ . The sign is that of $\frac{x - 1}{x}$ : positive for $x \lt 0$ and for $x \gt 1$ . Local minimum $f(1) = e$ .

As $x \to 0^-$ : $f'(x) = e^{1/x}\left(1 - \frac{1}{x}\right)$ ; with $t = 1/x \to -\infty$ , $e^t(1 - t) \to 0$ . The graph reaches the origin with a horizontal tangent.

5. Second derivative. $f''(x) = \dfrac{e^{1/x}}{x^3}$ : concave for $x \lt 0$ , convex for $x \gt 0$ . No inflection point (0 is excluded).

Graph of x times e to the 1 over x — The left branch rises along the oblique asymptote y equal to x plus 1, staying below it, and arrives flat at the origin. The right branch falls from plus infinity near the y axis, has its minimum at 1, and then rises again above the oblique asymptote.

Exercise 9. A logarithm with a split domain

Study $f(x) = \ln\left(x^2 - 3x + 2\right)$ .

Result: domain $x \lt 1$ or $x \gt 2$ , symmetric about the line $x = \frac{3}{2}$ ; vertical asymptotes $x = 1$ and $x = 2$ ; decreasing for $x \lt 1$ , increasing for $x \gt 2$ ; concave everywhere; zeros $\frac{3 \pm \sqrt 5}{2}$ .

Solution

1. Domain. $x^2 - 3x + 2 = (x - 1)(x - 2) \gt 0$ : $x \lt 1$ or $x \gt 2$ .

Symmetry. Substituting $3 - x$ for $x$ : $(3 - x)^2 - 3(3 - x) + 2 = x^2 - 3x + 2$ . So $f(3 - x) = f(x)$ : the graph is symmetric about the line $x = \frac{3}{2}$ , the axis of the parabola inside the logarithm.

2. Zeros and sign. $f = 0$ if $x^2 - 3x + 2 = 1$ , i.e. $x = \frac{3 \pm \sqrt 5}{2} \approx 0.382$ and $2.618$ . $f \gt 0$ if the argument exceeds 1: $x \lt 0.382$ or $x \gt 2.618$ .

3. Limits. $x \to 1^-$ and $x \to 2^+$ : the argument $\to 0^+$ , $f \to -\infty$ : vertical asymptotes. $x \to \pm\infty$ : $f \to +\infty$ , but $f(x)/x \to 0$ (the logarithm grows more slowly than $x$ ): no oblique asymptote.

4. First derivative. $f'(x) = \dfrac{2x - 3}{(x - 1)(x - 2)}$ . In the domain the denominator is positive: the sign is that of $2x - 3$ . For $x \lt 1$ it is negative (decreasing), for $x \gt 2$ positive (increasing). The zero $x = 3/2$ does not belong to the domain: no extrema.

5. Second derivative. $f''(x) = -\dfrac{2x^2 - 6x + 5}{(x - 1)^2(x - 2)^2}$ . The trinomial $2x^2 - 6x + 5$ has discriminant $36 - 40 \lt 0$ and is always positive: $f'' \lt 0$ , concave on both intervals.

Graph of the logarithm of x squared minus 3 x plus 2 — Two branches symmetric about the line x equal to three halves: the left one falls towards the asymptote x equal to 1, the right one rises from the asymptote x equal to 2. Both cross the x axis.

Exercise 10. A cusp

Study $f(x) = \sqrt[3]{x^2}\,(x - 5)$ .

Result: cusp with local maximum at $(0, 0)$ ; local minimum $\left(2, -3\sqrt[3]{4}\right) \approx (2, -4.76)$ ; inflection point $(-1, -6)$ ; no asymptotes.

Solution

1. Domain. $\mathbb R$ : the cube root is defined everywhere. Write $f(x) = x^{5/3} - 5x^{2/3}$ .

2. Zeros and sign. Zeros at 0 and 5. $\sqrt[3]{x^2} \ge 0$ , so the sign is that of $x - 5$ : negative for $x \lt 5$ (except at 0), positive after.

3. Limits. $\pm\infty$ at the ends; $f(x)/x \to +\infty$ : no asymptotes.

4. First derivative. For $x \ne 0$ :

$f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}\cdot\frac{x - 2}{\sqrt[3]{x}}$

Sign of $\frac{x - 2}{\sqrt[3]{x}}$ : positive for $x \lt 0$ (numerator and denominator negative) and for $x \gt 2$ ; negative between 0 and 2. Local minimum $f(2) = \sqrt[3]{4}\cdot(-3) \approx -4.76$ .

At 0: $\lim_{x\to 0^+}f'(x) = \frac{-2}{0^+} = -\infty$ and $\lim_{x\to 0^-}f'(x) = \frac{-2}{0^-} = +\infty$ . Infinite derivatives with opposite signs: cusp, which is also a local maximum (the function increases before and decreases after).

5. Second derivative. $f''(x) = \dfrac{10}{9}\cdot\dfrac{x + 1}{x^{4/3}}$ . The denominator $x^{4/3} = \sqrt[3]{x^4}$ is positive: concave for $x \lt -1$ , convex for $x \gt -1$ ( $x \ne 0$ ). Inflection point at $(-1, -6)$ .

Graph of the cube root of x squared times x minus 5 — The curve rises to the origin, where it forms an upward point, falls to the minimum at x equal to 2, rises again and crosses the x axis at 5. The inflection point is at minus 1.

Exercise 11. Exponential over linear

Study $f(x) = \dfrac{e^x}{x - 1}$ .

Result: asymptotes $x = 1$ and $y = 0$ (as $x \to -\infty$ ); local minimum $(2, e^2)$ ; concave for $x \lt 1$ , convex for $x \gt 1$ ; no inflection point.

Solution

1. Domain. $x \ne 1$ .

2. Zeros and sign. No zeros; sign of $x - 1$ . It meets the $y$ axis at $f(0) = -1$ .

3. Limits. $x \to 1^\pm$ : $f \to \pm\infty$ . $x \to -\infty$ : $e^x \to 0$ and the denominator $\to -\infty$ , $f \to 0^-$ : asymptote $y = 0$ on the left. $x \to +\infty$ : $f \to +\infty$ and $f(x)/x = \frac{e^x}{x(x - 1)} \to +\infty$ : no oblique asymptote.

4. First derivative. $f'(x) = \dfrac{e^x(x - 1) - e^x}{(x - 1)^2} = \dfrac{e^x(x - 2)}{(x - 1)^2}$ : negative for $x \lt 2$ ( $x \ne 1$ ), positive after. Local minimum $f(2) = e^2 \approx 7.39$ .

5. Second derivative. $f''(x) = \dfrac{e^x(x^2 - 4x + 5)}{(x - 1)^3}$ . $x^2 - 4x + 5$ has negative discriminant, so it is always positive: the sign is that of $x - 1$ . Concave to the left of 1, convex to the right, no inflection point.

Graph of e to the x over x minus 1 — On the left the graph starts from the x axis, passes through 0, minus 1 and falls towards minus infinity near x equal to 1. On the right it falls from plus infinity to the minimum at 2 and then rises quickly.

Exercise 12. How many solutions an equation has

As the real parameter $k$ varies, how many solutions does the equation $x\,e^{-x} = k$ have?

Result: none for $k \gt \frac{1}{e}$ ; one for $k = \frac{1}{e}$ ; two for $0 \lt k \lt \frac{1}{e}$ ; one for $k \le 0$ .

Solution

The solutions are the abscissas of the intersections between the graph of $f(x) = x\,e^{-x}$ and the horizontal line $y = k$ . The graph is the one studied in the lesson:

$f \to -\infty$ as $x \to -\infty$ , $f \to 0^+$ as $x \to +\infty$ ;
increasing on $(-\infty, 1)$ , decreasing on $(1, +\infty)$ , absolute maximum $f(1) = \frac{1}{e} \approx 0.368$ ;
$f \lt 0$ for $x \lt 0$ , $f \gt 0$ for $x \gt 0$ .

Move the line from top to bottom:

1. $k \gt \frac{1}{e}$ : the line lies above the maximum, no solutions.

2. $k = \frac{1}{e}$ : the line is tangent at the maximum, one solution, $x = 1$ .

3. $0 \lt k \lt \frac{1}{e}$ : two solutions, one in $(0, 1)$ on the increasing branch and one in $(1, +\infty)$ on the decreasing branch (which never reaches 0). For $k = 0.2$ they are about 0.259 and 2.543.

4. $k \le 0$ : the positive part of the graph no longer matters; the branch $x \le 0$ is increasing and goes from $-\infty$ to 0, so there is one solution, $x \le 0$ . For $k = -0.3$ it is about $-0.237$ .

The existence of the solutions follows from the intermediate value theorem (continuous function), uniqueness on each branch from monotonicity.

Graph of x times e to the minus x with four horizontal lines — Line k1, above the maximum, does not meet the graph; k2 touches it at the maximum; k3, between zero and the maximum, meets it twice; k4, negative, only once to the left of the origin.

0 Comments

Oldest

Newest Most Voted

Cookie	Duration	Description
cookielawinfo-checkbox-analytics	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Analytics".
cookielawinfo-checkbox-functional	11 months	The cookie is set by GDPR cookie consent to record the user consent for the cookies in the category "Functional".
cookielawinfo-checkbox-necessary	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookies is used to store the user consent for the cookies in the category "Necessary".
cookielawinfo-checkbox-others	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Other.
cookielawinfo-checkbox-performance	11 months	This cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Performance".
viewed_cookie_policy	11 months	The cookie is set by the GDPR Cookie Consent plugin and is used to store whether or not user has consented to the use of cookies. It does not store any personal data.

Basic level

Exercise 1. A cubic polynomial

Exercise 2. A rational function with a horizontal asymptote

Exercise 3. The Gaussian

Intermediate level

Exercise 4. Vertical and oblique asymptotes

Exercise 5. Logarithm over x

Exercise 6. A root with vertical tangents

Exercise 7. An absolute value and a corner point

Exam level

Exercise 8. An asymptote on one side only

Exercise 9. A logarithm with a split domain

Exercise 10. A cusp

Exercise 11. Exponential over linear

Exercise 12. How many solutions an equation has

Related articles

China’s new rules for assisted and automated driving (GB 47955 and GB 44721)

The chips and sensors behind China’s assisted driving

Remove the MacBook screen so it can work as a desktop computer

Robotaxis in China in 2026: Baidu Apollo Go, Pony.ai and WeRide

Root locus in depth: the rules proved, difficult cases, MATLAB and Python

Navigate on Autopilot in China: how city NOA became a mass-market feature