Sketching Graphs by Function Type · Mathematics AP Calculus AB

The sketching checklist

Every graph study follows the same order. Each step uses information from the earlier ones, so don't skip around.

Domain. Where is $f$ defined?
Symmetry. Is $f$ even or odd?
Intercepts. Find $f(0)$ and solve $f(x) = 0$ when you can.
Asymptotes. Vertical, horizontal and slant (oblique), using limits.
First derivative. Find where $f$ is increasing or decreasing and locate the relative extrema.
Second derivative. Find where $f$ is concave up or concave down and locate the points of inflection.
Sketch. Draw asymptotes as dashed lines, plot the key points, and connect them using the shape each interval requires.

Symmetry halves your work: study $x \ge 0$ and reflect.

Where a domain can break

If $f$ contains…	Then require…
a denominator $g(x)$	$g(x) \ne 0$
an even root $\sqrt{g(x)}$	$g(x) \ge 0$
a logarithm $\ln g(x)$	$g(x) > 0$
$e^{g(x)}$, a polynomial, $\sin$, $\cos$	no restriction by itself (restrict only what is inside $g$)

Practice★☆☆

Determine the parity of the following functions, justifying mathematically whether they are even ($f(x) = f(-x)$), odd ($f(-x) = -f(x)$), or neither: a)…

Determine the parity of the following functions, justifying mathematically whether they are even ($f(x) = f(-x)$), odd ($f(-x) = -f(x)$), or neither:

a) $f(x) = x^2 \cdot \cos(x)$
b) $g(x) = \frac{x^3}{x^2 + 1}$
c) $h(x) = e^x - x$

Solve

Polynomial functions: no asymptotes, everything lives in $f'$ and $f''$

Full study of $f(x) = x^3 - 3x^2 + 4$

Domain $(-\infty, \infty)$. The function mixes odd and even powers with a constant, so it is neither even nor odd.

Intercepts. $f(0) = 4$. For $f(x) = 0$, try $x = -1$: $-1 - 3 + 4 = 0$ ✓. Dividing out $(x+1)$ gives $f(x) = (x+1)(x-2)^2$, so the $x$-intercepts are $x = -1$ and $x = 2$ (a double zero, where the graph touches the axis).

First derivative. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so the critical numbers are $0$ and $2$. The sign of $f'$ is $+, -, +$. So $f$ is increasing on $(-\infty, 0)$ and $(2, \infty)$, and decreasing on $(0, 2)$. By the First Derivative Test there is a relative maximum at $(0, 4)$ and a relative minimum at $(2, 0)$.

Second derivative. Find $f''$, its zero, and the concavity. Try it before you look.

$f''(x) = 6x - 6$ is $0$ at $x = 1$. It is negative for $x < 1$ and positive for $x > 1$. So $f$ is concave down on $(-\infty, 1)$ and concave up on $(1, \infty)$, with a point of inflection at $(1, 2)$.

Practice★☆☆

Given the polynomial function $f(x) = x^3 - 3x^2 + 4$, carry out a basic study of its behavior following these steps: a) Calculate the points of intersection with the…

Given the polynomial function $f(x) = x^3 - 3x^2 + 4$, carry out a basic study of its behavior following these steps:

a) Calculate the points of intersection with the coordinate axes.
b) Find the increasing and decreasing intervals using the first derivative $f'(x)$.
c) Determine the $(x, y)$-coordinates of its relative maxima and minima.

Solve

Practice★☆☆

Given the function $f(x) = x^4 - 2x^2$, study its curvature by carrying out the following steps: a) Calculate the second derivative $f''(x)$. b) Find the intervals of…

Given the function $f(x) = x^4 - 2x^2$, study its curvature by carrying out the following steps:

a) Calculate the second derivative $f''(x)$.
b) Find the intervals of concavity ($\cup$) and convexity ($\cap$).
c) Find the points of inflection of the function.

Solve

Practice★★☆

Determine the values of the parameters $a$ and $b$ so that the function $f(x) = ax^3 + bx^2 - x + 4$ has a point of inflection at $x = 1$ and the slope of the tangent…

Determine the values of the parameters $a$ and $b$ so that the function $f(x) = ax^3 + bx^2 - x + 4$ has a point of inflection at $x = 1$ and the slope of the tangent line at that point is $1$. Once the parameters have been found:

Study the monotonicity of the resulting function.
Determine if it has any relative maximum or minimum.
Justify whether the function has asymptotes.

Solve

Rational functions: reading the asymptotes

Rational functions use every tool on the checklist: a restricted domain and up to three kinds of asymptotes.

Asymptotes of $f(x) = \dfrac{p(x)}{q(x)}$

Vertical: $x = c$ where $q(c) = 0$ but $p(c) \ne 0$. At least one one-sided limit is $\pm\infty$. If a factor cancels, you get a hole, not an asymptote.
Horizontal (limit at infinity is a finite number $L$):

Degrees	Result
$\deg p < \deg q$	$y = 0$
$\deg p = \deg q$	$y = $ ratio of leading coefficients
$\deg p = \deg q + 1$	no horizontal; there is a slant asymptote $y = mx + b$
$\deg p \ge \deg q + 2$	neither

Slant: divide $p$ by $q$. The quotient $mx + b$ is the asymptote, and the remainder term goes to $0$.

Why? Divide top and bottom by the highest power of $x$ in the denominator. Terms with $x$ in the denominator vanish at infinity, so only the leading terms matter.

Asymptotes of $f(x) = \dfrac{x^2 - 2x + 3}{x + 1}$

Vertical. The denominator is $0$ at $x = -1$, and the numerator there is $1 + 2 + 3 = 6 \ne 0$. So $x = -1$ is a vertical asymptote.

Slant. The numerator has degree 2 and the denominator degree 1, so divide: $x^2 - 2x + 3 = (x+1)(x-3) + 6$. Therefore

$$f(x) = x - 3 + \frac{6}{x+1}.$$

Since $\dfrac{6}{x+1} \to 0$ as $x \to \pm\infty$, the line $y = x - 3$ is a slant asymptote.

Position. The remainder $\dfrac{6}{x+1}$ is positive for $x > -1$, so the curve is above the slant asymptote as $x \to +\infty$. It is negative for $x < -1$, so the curve is below it as $x \to -\infty$.

Behavior at the vertical asymptote. Find $\displaystyle\lim_{x\to -1^+} f(x)$ and $\displaystyle\lim_{x\to -1^-} f(x)$.

The numerator is near $6 > 0$. For $x \to -1^+$ the denominator is a small positive number, so $f \to +\infty$. For $x \to -1^-$ it is a small negative number, so $f \to -\infty$.

When a problem gives a model like $P(t) = \dfrac{40t}{t+2}$, the horizontal asymptote has a meaning: $\lim_{t\to\infty} P(t)$ is the long-run value the quantity approaches but never has to reach.

Practice★☆☆

Consider the rational function $f(x) = \frac{2x + 4}{x - 3}$. To understand its structure before representing it, analyze its main elements: a) Determine its domain of…

Consider the rational function $f(x) = \frac{2x + 4}{x - 3}$. To understand its structure before representing it, analyze its main elements:

a) Determine its domain of definition.
b) Calculate the equations of its vertical and horizontal asymptotes.
c) Study the symmetry of the function. Is it even, odd, or does it have no symmetry?

Solve

Practice★☆☆

Observe the following graph of a function $f(x)$ and extract the requested information: a) State the values of $x$ where the vertical asymptotes are located. b) Estimate…

Observe the following graph of a function $f(x)$ and extract the requested information:

a) State the values of $x$ where the vertical asymptotes are located.
b) Estimate the intervals where the function is increasing.
c) Identify the value of the $\lim_{x \to \infty} f(x)$ by observing the behavior of the graph.

Solve

Practice★☆☆

A biologist studies the evolution of a bacteria population in a controlled environment. The number of bacteria (in thousands) is given by the function…

A biologist studies the evolution of a bacteria population in a controlled environment. The number of bacteria (in thousands) is given by the function $P(t) = \frac{40t}{t + 2}$, where $t$ is the time in hours since the start of the experiment ($t \geq 0$).

a) What is the population at the start of the experiment ($t = 0$)?
b) What is the limit of the population as time increases indefinitely? Interpret the result in terms of asymptotes.
c) Does the population increase or decrease over time? Justify your answer using the derivative $P'(t)$.

Solve

Rational functions: a full sketch

Sketch $f(x) = \dfrac{x^2}{x^2 - 4}$

Domain and symmetry. Domain: all reals except $x = \pm 2$. Since $f(-x) = f(x)$, the function is even, so the graph is symmetric about the $y$-axis.

Intercepts and asymptotes. $f(0) = 0$, and $x^2 = 0$ only at $0$, so the only intercept is the origin. At $x = \pm 2$ the numerator is $4 \ne 0$, so $x = 2$ and $x = -2$ are vertical asymptotes. The degrees are equal, so the horizontal asymptote is $y = \dfrac{1}{1} = 1$.

Position relative to the asymptotes. $f(x) - 1 = \dfrac{4}{x^2 - 4}$, which is positive for $|x| > 2$ (curve above $y = 1$) and negative for $|x| < 2$. Near the vertical asymptotes: $f \to +\infty$ as $x \to 2^+$ and $f \to -\infty$ as $x \to 2^-$. By symmetry, $f \to +\infty$ as $x \to -2^-$ and $f \to -\infty$ as $x \to -2^+$.

First derivative. $f'(x) = \dfrac{2x(x^2-4) - x^2 \cdot 2x}{(x^2-4)^2} = \dfrac{-8x}{(x^2-4)^2}$. The denominator is positive, so the sign is the sign of $-8x$. Thus $f$ is increasing on $(-\infty, -2)$ and $(-2, 0)$, and decreasing on $(0, 2)$ and $(2, \infty)$. Since $f'$ changes from positive to negative at $x = 0$, there is a relative maximum at $(0, 0)$.

Second derivative. Compute $f''$ and decide about inflection points. Try it first.

$f''(x) = \dfrac{8(3x^2 + 4)}{(x^2 - 4)^3}$. The numerator is always positive, so the sign is the sign of $x^2 - 4$. The graph is concave up on $(-\infty, -2)$ and $(2, \infty)$, and concave down on $(-2, 2)$. There is no point of inflection: $f''$ is never $0$, and it changes sign only at $x = \pm 2$, which are not in the domain.

Back to the stream: Emma's viewers

Sketch $V(t) = \dfrac{30t}{t^2 + 9}$ for $t \ge 0$.

$V(0) = 0$ and $V(t) > 0$ for $t > 0$. The degree of the numerator (1) is less than that of the denominator (2), so $\lim_{t\to\infty} V(t) = 0$: the horizontal asymptote is $V = 0$.

$V'(t) = \dfrac{30(t^2+9) - 30t \cdot 2t}{(t^2+9)^2} = \dfrac{30(9 - t^2)}{(t^2+9)^2}$. For $t \ge 0$ this is positive on $(0, 3)$ and negative on $(3, \infty)$. By the First Derivative Test, $V$ has a relative (and absolute) maximum at $t = 3$ with $V(3) = \dfrac{90}{18} = 5$: the audience peaks at 5 thousand viewers per hour, 3 hours in.

Concavity. For $t > 0$, find where the audience stops dropping faster and faster.

$V''(t) = \dfrac{60t(t^2 - 27)}{(t^2+9)^3}$. For $t > 0$ it is negative on $(0, 3\sqrt3)$ and positive on $(3\sqrt3, \infty)$, with $3\sqrt 3 \approx 5.2$. So there is a point of inflection near $t \approx 5.2$ hours, where $V \approx 4.33$. After that the decline keeps going but flattens toward $0$.

Practice★★☆

Given the rational function $f(x) = \frac{2x^2}{x^2 - 9}$, carry out a complete study for its graphical representation following these steps: 1. Find the domain of…

Given the rational function $f(x) = \frac{2x^2}{x^2 - 9}$, carry out a complete study for its graphical representation following these steps:

Find the domain of definition and the points of intersection with the coordinate axes.
Study the existence of symmetries in the function.
Determine the vertical and horizontal asymptotes and the position of the curve with respect to them.
Calculate the increasing and decreasing intervals, and determine its relative maxima and minima.
Sketch the graph of the function based on the previous results.

Solve

Practice★★☆

Consider the function $f(x) = \frac{x^2 + 3}{x - 1}$. Answer the following: 1. Find the equations of its asymptotes. If an oblique asymptote exists, determine the…

Consider the function $f(x) = \frac{x^2 + 3}{x - 1}$. Answer the following:

Find the equations of its asymptotes. If an oblique asymptote exists, determine the position of the curve with respect to it when $x \to +\infty$ and $x \to -\infty$.
Find the intervals of concavity and convexity, and check if there are any points of inflection.
Use the information on the increase and extrema to draw an accurate sketch of the graph.

Solve

Practice★★☆

A biological study on an insect population in a controlled environment follows the function $P(t) = \frac{50t}{t^2 + 4} + 2$, where $P$ is the population in thousands…

A biological study on an insect population in a controlled environment follows the function $P(t) = \frac{50t}{t^2 + 4} + 2$, where $P$ is the population in thousands and $t$ is the time in months since the start of the experiment ($t \geq 0$).

Determine at what time the population reaches its maximum value and what this value is.
What value does the population tend to if the experiment is prolonged indefinitely? Justify your answer by calculating limits.
Study the curvature of the function to identify at what time the population decline begins to slow down (point of inflection).

Solve

Absolute value functions: split into pieces before differentiating

Sketch $f(x) = |x^2 - 2x|$

Let $g(x) = x^2 - 2x = x(x-2)$. It is $0$ at $x = 0$ and $x = 2$, positive for $x < 0$ and $x > 2$, and negative on $(0, 2)$.

$$f(x) = \begin{cases} x^2 - 2x & \text{if } x \le 0 \text{ or } x \ge 2 \\ -(x^2 - 2x) & \text{if } 0 < x < 2 \end{cases}$$

That is the parabola with its middle part flipped above the $x$-axis.

Derivatives piece by piece. On $x < 0$ and $x > 2$, $f'(x) = 2x - 2$. On $(0, 2)$, $f'(x) = -2x + 2$. At $x = 0$ the left slope is $-2$ and the right slope is $+2$. At $x = 2$ the left slope is $-2$ and the right slope is $2$. The one-sided derivatives differ, so $f$ is not differentiable at $0$ or $2$ (corners).

Extrema. Use the sign of $f'$ to find them.

$f'$ is negative on $(-\infty, 0)$, positive on $(0, 1)$, negative on $(1, 2)$ and positive on $(2, \infty)$. So there are relative minima at $(0, 0)$ and $(2, 0)$, which are corners where $f'$ doesn't exist, and a relative maximum at $(1, 1)$, where $f'(1) = 0$.

Practice★☆☆

Absolute value functions often have points where the graph changes direction abruptly. Let the function be $f(x) = |x - 4|$. a) Express the function as a piecewise…

Absolute value functions often have points where the graph changes direction abruptly. Let the function be $f(x) = |x - 4|$.

a) Express the function as a piecewise function.
b) Indicate the point on the graph where the function is continuous but not differentiable (corner point).
c) Sketch its graph indicating the vertex.

Solve

Practice★★☆

Given the absolute value function $f(x) = |x^2 - 4x + 3|$: 1. Express the function as a piecewise function. 2. Study the differentiability of the function at the points…

Given the absolute value function $f(x) = |x^2 - 4x + 3|$:

Express the function as a piecewise function.
Study the differentiability of the function at the points $x = 1$ and $x = 3$. What happens geometrically at these points?
Find the relative extrema of the function and represent it graphically.

Solve

Practice★★★

Given the function: $$f(x) = \frac{x \cdot |x|}{x - 1}$$ 1. Express the function as a piecewise function by removing the modulus. 2. Study the continuity and…

Given the function:

$$f(x) = \frac{x \cdot |x|}{x - 1}$$

Express the function as a piecewise function by removing the modulus.
Study the continuity and differentiability of the function over its entire domain, paying special attention to the point $x = 0$.
Find the asymptotes of the graph of $f$.
Determine the increasing and decreasing intervals and the relative extrema.
Sketch the graph of the function, clearly indicating the points where the derivative does not exist, if any.

Solve

Exponential and logarithmic functions

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

Domain: all reals. Intercepts: $f(0) = 0$, and $x\,e^{-x} = 0$ only when $x = 0$ (the factor $e^{-x}$ is never $0$).

Derivatives. $f'(x) = e^{-x} - x e^{-x} = (1-x)e^{-x}$. Since $e^{-x} > 0$, the sign is the sign of $1 - x$: increasing on $(-\infty, 1)$, decreasing on $(1, \infty)$. There is a relative maximum at $\left(1, \tfrac{1}{e}\right) \approx (1, 0.37)$.

$f''(x) = -e^{-x} - e^{-x} + x e^{-x} = (x - 2)e^{-x}$. It is negative for $x < 2$ and positive for $x > 2$. The graph is concave down on $(-\infty, 2)$, concave up on $(2, \infty)$, with a point of inflection at $(2, 2e^{-2}) \approx (2, 0.27)$.

Limits at infinity. Which end has an asymptote?

As $x \to +\infty$, $x e^{-x} = \dfrac{x}{e^x} \to 0$ (exponentials beat polynomials), so $y = 0$ is a horizontal asymptote on the right. As $x \to -\infty$, $x \to -\infty$ and $e^{-x} \to \infty$, so $f \to -\infty$: no asymptote on the left.

Practice★☆☆

The domain of a function is fundamental for its representation. Find the domain of the following transcendental functions: a) $f(x) = \ln(x^2 - 1)$ b)…

The domain of a function is fundamental for its representation. Find the domain of the following transcendental functions:

a) $f(x) = \ln(x^2 - 1)$
b) $g(x) = e^{\frac{1}{x}}$
c) $h(x) = \sqrt{x + 5}$

Briefly explain why there are restrictions in each case.

Solve

Practice★★☆

Let the function be $f(x) = (x - 2)e^x$. Carry out a detailed study of it to represent it graphically, including: 1. Domain and points of intersection with the axes. 2.…

Let the function be $f(x) = (x - 2)e^x$. Carry out a detailed study of it to represent it graphically, including:

Domain and points of intersection with the axes.
Horizontal asymptotes and behavior at infinity.
Intervals of monotonicity and relative extrema.
Points of inflection and curvature.

Solve

Practice★★☆

Study and represent the logarithmic function $f(x) = \ln(x^2 + 1)$. In your analysis you must include: 1. Proof that the function is even. How does this affect its…

Study and represent the logarithmic function $f(x) = \ln(x^2 + 1)$. In your analysis you must include:

Proof that the function is even. How does this affect its representation?
Determination of the increasing and decreasing intervals.
Study of the curvature and search for points of inflection.
Does the function have asymptotes of any kind? Justify your answer.

Solve

Piecewise functions: continuity first, then each piece

Practice★★☆

Consider the piecewise function: $$f(x) = \begin{cases} \frac{1}{x-2} & \text{if } x < 0 \\ x^2 - 2x - \frac{1}{2} & \text{if } x \geq 0 \end{cases}$$ 1. Study the…

Consider the piecewise function:

$$f(x) = \begin{cases} \frac{1}{x-2} & \text{if } x < 0 \\ x^2 - 2x - \frac{1}{2} & \text{if } x \geq 0 \end{cases}$$

Study the continuity and differentiability of the function at $x = 0$.
Find the asymptotes of the function in its domain of definition.
Calculate the local minima and maxima of each piece in its corresponding interval.
Sketch the graph of the function.

Solve

Practice★★★

Consider the piecewise function: $$f(x) = \begin{cases} \dfrac{x^2 - 4}{x+2} & \text{if } x < -2 \\ a \cdot e^{x+2} + b & \text{if } -2 \le x < 0 \\ \dfrac{x+1}{x^2+1} &…

Consider the piecewise function:

$$f(x) = \begin{cases} \dfrac{x^2 - 4}{x+2} & \text{if } x < -2 \\ a \cdot e^{x+2} + b & \text{if } -2 \le x < 0 \\ \dfrac{x+1}{x^2+1} & \text{if } x \ge 0 \end{cases}$$

Carry out a complete study of the function following these steps:

Find the values of $a$ and $b$ so that the function is continuous at $x = -2$ and has a relative extremum at $x = -1$.
For the values found, study the differentiability at $x = -2$ and $x = 0$.
Calculate the asymptotes of the function over its entire domain.
Determine the increasing and decreasing intervals, as well as the relative maxima and minima in the interval $(-\infty, 0)$.

Solve

Practice★★★

A ramp profile for a skatepark is to be designed. The side profile follows the curve of the function: $$f(x) = \frac{\ln(x^2+1)}{x}$$ for $x \in [-5, 5], x \neq 0$.…

A ramp profile for a skatepark is to be designed. The side profile follows the curve of the function:

$$f(x) = \frac{\ln(x^2+1)}{x}$$

for $x \in [-5, 5], x \neq 0$. Answer the following:

Show that the function is odd. What does this imply for its graphical representation?
Define the value of $f(0)$ so that the function is continuous at $x=0$. To do this, calculate $\lim_{x \to 0} f(x)$ (you can use L'Hôpital's rule).
Is the function differentiable at $x=0$ with the previous definition? Justify your answer.
Determine the points on the ramp where the gradient is zero.
Study whether there are horizontal asymptotes when the domain is extended to the entire real line.

Solve

Mixed functions: when the tools combine

Some functions blend families: a product of $x$ and a logarithm, a logarithm of an absolute value minus a line, an exponential with $\tfrac1x$ in the exponent. First identify the pieces, then use the shortcut of each piece.

Study $f(x) = x\ln|x|$

Domain and symmetry. Domain: $x \ne 0$. Since $f(-x) = (-x)\ln|-x| = -x\ln|x| = -f(x)$, $f$ is odd. Intercepts: $\ln|x| = 0$ gives $x = \pm 1$. There is no $y$-intercept.

Behavior near $0$. $\displaystyle\lim_{x\to0^+} x\ln x$ is $0 \cdot (-\infty)$. Rewrite: $\displaystyle\lim_{x\to0^+}\frac{\ln x}{1/x} = \lim_{x\to0^+}\frac{1/x}{-1/x^2} = \lim_{x\to0^+}(-x) = 0$. By oddness the left limit is $0$ too. So there is no vertical asymptote, only a hole at $(0, 0)$.

First derivative. $f'(x) = \ln|x| + 1$, which is $0$ when $|x| = \tfrac1e$. For $x > 0$: negative on $\left(0, \tfrac1e\right)$, positive on $\left(\tfrac1e, \infty\right)$, so there is a relative minimum at $\left(\tfrac1e, -\tfrac1e\right)$. By symmetry there is a relative maximum at $\left(-\tfrac1e, \tfrac1e\right)$.

Concavity and end behavior. Work them out before checking.

$f''(x) = \dfrac1x$: concave down on $(-\infty, 0)$, concave up on $(0, \infty)$. There is no inflection point because $x = 0$ is not in the domain. As $x \to \infty$, $f \to \infty$, and $\dfrac{f(x)}{x} = \ln x \to \infty$, so there is no slant asymptote either.

Practice★★★

A communications satellite follows a trajectory whose signal intensity received on Earth is given by the function: $$S(t) = \ln |t^2 - 9| - \frac{t}{2}$$ where $t$…

A communications satellite follows a trajectory whose signal intensity received on Earth is given by the function:

$$S(t) = \ln |t^2 - 9| - \frac{t}{2}$$

where $t$ represents the time in hours since the start of the tracking ($t \ge 0$).

Determine the domain of the function and identify the instants when the signal is completely interrupted.
Find the vertical asymptotes and explain their meaning in this context.
Calculate the increasing and decreasing intervals of the signal intensity.
Determine at what time the signal reaches its relative maximum value and sketch the graph of the intensity trajectory.

Solve

Practice★★★

Consider the function $f(x) = (x-2)^2 \cdot e^x$. 1. Calculate the points of intersection with the axes and the asymptotes of the function. 2. Determine the intervals of…

Consider the function $f(x) = (x-2)^2 \cdot e^x$.

Calculate the points of intersection with the axes and the asymptotes of the function.
Determine the intervals of monotonicity and the local extrema (maxima and minima).
Find the intervals of curvature and the points of inflection.
Knowing that $e^{-2} \approx 0.135$ and $e^0 = 1$, sketch the graph of the function.
Calculate the area of the triangle formed by the origin of coordinates and the points on the graph where the tangent is horizontal.

Solve

Practice★★★

Study and graphically represent the "mixed type" function: $$f(x) = x \cdot e^{1/x}$$ To do this, analyze in detail: 1. The domain of definition and the behavior of the…

Study and graphically represent the "mixed type" function:

$$f(x) = x \cdot e^{1/x}$$

To do this, analyze in detail:

The domain of definition and the behavior of the function near $x = 0$ by calculating the one-sided limits $\lim_{x \to 0^+} f(x)$ and $\lim_{x \to 0^-} f(x)$.
The vertical and oblique asymptotes.
The existence of relative extrema using the first derivative.
The concavity and convexity through the study of the sign of $f''(x)$.

Solve

Working backwards: parameters and sketches from conditions

Many problems give you information about a graph and ask you to find the formula or draw the curve. The key is to translate each condition into an equation, or into a shape.

To draw from conditions about $f'$ and $f''$, remember that on each interval the signs of $f'$ and $f''$ give one of four shapes.

Practice★★★

Let the rational function be defined by: $$f(x) = \frac{x^3 + ax^2 + b}{x^2 - 1}$$ Determine the values of the real parameters $a$ and $b$ knowing that the line…

Let the rational function be defined by:

$$f(x) = \frac{x^3 + ax^2 + b}{x^2 - 1}$$

Determine the values of the real parameters $a$ and $b$ knowing that the line $y = x + 2$ is an oblique asymptote of the function. Once the parameters have been calculated:

Find the domain and the vertical asymptotes, studying the position of the curve with respect to them.
Find the points of intersection with the axes.
Study the existence of points of inflection. Is it possible that this function has no points of inflection in its domain? Justify your answer.

Solve

Practice★★★

Draw a sketch of the graph of a continuous function $f(x)$ that simultaneously satisfies the following eight conditions: Its domain is $\mathbb{R} - \{1\}$. It has a…

Draw a sketch of the graph of a continuous function $f(x)$ that simultaneously satisfies the following eight conditions:

Its domain is $\mathbb{R} - \{1\}$.
It has a vertical asymptote at $x = 1$ such that $\lim_{x \to 1^-} f(x) = +\infty$ and $\lim_{x \to 1^+} f(x) = -\infty$.
It has an oblique asymptote $y = x$ when $x \to +\infty$.
It has a horizontal asymptote $y = -2$ when $x \to -\infty$.
$f'(x) < 0$ in the interval $(-\infty, -2)$ and $f'(x) > 0$ in $(-2, 1) \cup (1, +\infty)$.
It has a relative minimum at the point $(-2, -4)$.
$f''(x) > 0$ in $(-\infty, 1)$ and $f''(x) < 0$ in $(1, +\infty)$.
It intersects the $OX$-axis only at the points $x = -4$ and $x = 2$.

Solve

Back to the stream: Emma's audience peaks at $t = 3$ hours with 5 thousand viewers per hour, and since the horizontal asymptote is $V = 0$, it fades toward zero without ever being negative. You found that from $V'$ and a limit, with no calculator.

Summary

Checklist: domain → symmetry → intercepts → asymptotes → $f'$ → $f''$ → sketch. Identify the type first, since each type lets you skip some checks.
Polynomials: domain $(-\infty, \infty)$, no asymptotes; everything comes from intercepts, end behavior, $f'$ and $f''$.
Rational functions: vertical asymptotes where the denominator is $0$ (after canceling holes); horizontal by comparing degrees; slant when the degree is one more (divide). A curve may cross its horizontal or slant asymptote, never a vertical one.
Absolute value: split into pieces by the sign of the inside; corners where the inside is $0$ with nonzero slope; critical numbers include points where $f'$ is undefined.
Exponentials and logs: $e^{g} > 0$ (no $x$-intercepts), usually a horizontal asymptote on one side; $\ln g$ needs $g > 0$ and has a vertical asymptote where $g \to 0^+$.
Piecewise: check continuity first, then compare one-sided derivatives. Rewrite $0\cdot\infty$ forms as quotients and use L'Hôpital's Rule.
Justify with AP language: First Derivative Test for extrema, sign change of $f''$ for inflection.

Let's practice! 8 medium exercises in this section

★★☆ Given the rational function $f(x) = \frac{2x^2}{x^2 - 9}$, carry out a complete study for…★★☆ Consider the function $f(x) = \frac{x^2 + 3}{x - 1}$. Answer the following: 1. Find the…★★☆ A biological study on an insect population in a controlled environment follows the…★★☆ Given the absolute value function $f(x) = |x^2 - 4x + 3|$: 1. Express the function as a…

Let's practice! 8 hard exercises in this section

★★★ Consider the piecewise function: $$f(x) = \begin{cases} \dfrac{x^2 - 4}{x+2} & \text{if }…★★★ Let the rational function be defined by: $$f(x) = \frac{x^3 + ax^2 + b}{x^2 - 1}$$…★★★ A communications satellite follows a trajectory whose signal intensity received on Earth…★★★ Given the function: $$f(x) = \frac{x \cdot |x|}{x - 1}$$ 1. Express the function as a…