Graphs of functions: shapes, intercepts and asymptotes

A graph of a function turns an algebraic rule into a picture. A good sketch shows much more than a collection of plotted points: it shows where the function is positive or negative, where it increases or decreases, whether it crosses an axis, and how it behaves for very large values of $|x|$.

This lesson develops the graph recognition and sketching skills used throughout A-level Mathematics. It focuses on extracting the important features from an equation without plotting hundreds of points.

Before you begin

You should be comfortable with:

• function notation, including the meanings of $f(x)$, domain and range;
• solving linear and quadratic equations;
• factorising polynomials;
• indices and logarithms;
• coordinates and the equations of straight lines.

The information carried by a graph

The graph of $y=f(x)$ is the set of all points $(x,f(x))$ for allowed values of $x$. When sketching it, look for the following features.

Feature	Algebraic question	Meaning on the graph
$y$-intercept	What is $f(0)$?	Where the graph meets the $y$-axis
$x$-intercepts	Where is $f(x)=0$?	The roots, where the graph meets the $x$-axis
Sign	Where is $f(x)>0$ or $f(x)<0$?	Whether the graph is above or below the $x$-axis
Stationary points	Where is $f'(x)=0$?	Usually a local maximum, local minimum or stationary point of inflection
Asymptotes	Which lines does the graph approach?	Long-term or near-discontinuity behaviour
End behaviour	What happens as $x\to\infty$ and $x\to-\infty$?	The directions of the ends of the graph
Symmetry	Is $f(-x)=f(x)$ or $f(-x)=-f(x)$?	Symmetry about the $y$-axis or the origin

A sketch does not need a scale unless the question asks for one, but every known intercept, turning point and asymptote should be labelled.

A reliable sketching method

For most functions, use this order:

1. State any restrictions on the domain.
2. Find the intercepts.
3. Identify asymptotes or discontinuities.
4. Determine symmetry, if present.
5. Find stationary points if the necessary calculus is available.
6. Determine end behaviour.
7. Join the features with a shape consistent with the function.

The final step matters. A graph is not made by joining special points with straight line segments. Those points constrain the curve, but the equation determines its shape between them.

Polynomial graphs

A polynomial has the form

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$

where $n$ is a non-negative integer. Polynomial graphs are continuous and smooth everywhere. They have no breaks, holes or vertical asymptotes.

End behaviour

For large $|x|$, the highest power term $a_nx^n$ dominates. Lower power terms become comparatively insignificant.

Degree and leading coefficient	As $x\to-\infty$	As $x\to\infty$
Even degree, $a_n>0$	$f(x)\to\infty$	$f(x)\to\infty$
Even degree, $a_n<0$	$f(x)\to-\infty$	$f(x)\to-\infty$
Odd degree, $a_n>0$	$f(x)\to-\infty$	$f(x)\to\infty$
Odd degree, $a_n<0$	$f(x)\to\infty$	$f(x)\to-\infty$

For example, $-2x^4+7x-1$ has both ends pointing down because its leading term is $-2x^4$. You do not need to know its roots to determine this.

Roots and repeated roots

Suppose a polynomial contains a factor $(x-a)^m$. Then $x=a$ is a root of multiplicity $m$.

• If $m$ is odd, the graph crosses the $x$-axis at $x=a$.
• If $m$ is even, the graph touches the $x$-axis and turns around at $x=a$.

For higher multiplicities, the graph becomes flatter near the root. The distinction between crossing and touching is determined by whether the multiplicity is odd or even.

Worked example 1: sketch a cubic from its factors

Sketch

$$y=(x+2)(x-1)^2.$$

Step 1: find the roots.

$$(x+2)(x-1)^2=0$$

gives $x=-2$ and $x=1$.

The root $x=-2$ has multiplicity $1$, so the graph crosses there. The root $x=1$ has multiplicity $2$, so the graph touches the axis and turns there.

Step 2: find the $y$-intercept.

$$y=(0+2)(0-1)^2=2.$$

The graph passes through $(0,2)$.

Step 3: determine the ends.

The leading term is $x\cdot x^2=x^3$. Therefore

$$y\to-\infty\text{ as }x\to-\infty, \qquad y\to\infty\text{ as }x\to\infty.$$

Step 4: combine the information.

The curve rises from the bottom left, crosses at $(-2,0)$, passes through $(0,2)$, falls to touch at $(1,0)$, then rises towards the top right.

Notice that the factorised form reveals the roots and their behaviour immediately. Expanding first would hide the most useful information.

Check your understanding

Without expanding, describe the key features of

$$y=-(x+1)^2(x-3).$$

Answer

The roots are $x=-1$ and $x=3$. The graph touches the axis at $x=-1$ because that root has even multiplicity, and crosses at $x=3$ because that root has odd multiplicity. The $y$-intercept is

$$y=-(1)^2(-3)=3.$$

The leading term is $-x^3$, so the graph rises to the left and falls to the right.

Reciprocal and rational graphs

The basic reciprocal function is

$$y=\frac{1}{x}.$$

It is undefined at $x=0$, so $x=0$ is a vertical asymptote. Also,

$$\frac{1}{x}\to0$$

as $x\to\infty$ or $x\to-\infty$, so $y=0$ is a horizontal asymptote. The graph has one branch in quadrant I and one in quadrant III.

An asymptote is a line that the graph approaches. It is not automatically a line that the graph can never cross. For example, a rational graph can cross a horizontal asymptote at a finite value of $x$.

The form $\boldsymbol{\frac{a}{x-b}+c}$

For

$$y=\frac{a}{x-b}+c,$$

the asymptotes are

$$x=b \qquad\text{and}\qquad y=c.$$

Their intersection $(b,c)$ is the centre of the graph. If $a>0$, the branches lie above-right and below-left of the centre. If $a<0$, they lie above-left and below-right.

Worked example 2: a translated reciprocal graph

Sketch

$$y=\frac{6}{x-2}-1.$$

Asymptotes:

$$x=2, \qquad y=-1.$$

$y$-intercept: set $x=0$.

$$y=\frac{6}{-2}-1=-4,$$

so the graph passes through $(0,-4)$.

$x$-intercept: set $y=0$.

$$0=\frac{6}{x-2}-1$$ $$1=\frac{6}{x-2}$$ $$x-2=6$$ $$x=8.$$

Since $6>0$, the branches occupy the above-right and below-left regions relative to the centre $(2,-1)$. Draw the asymptotes first as dashed lines, plot $(0,-4)$ and $(8,0)$, then draw two smooth branches approaching the asymptotes.

Finding asymptotes from a rational expression

Consider

$$f(x)=\frac{2x+1}{x-3}.$$

The denominator is zero at $x=3$, while the numerator is not, so $x=3$ is a vertical asymptote. Divide or rearrange:

$$2x+1=2(x-3)+7,$$

so

$$f(x)=2+\frac{7}{x-3}.$$

Therefore $y=2$ is the horizontal asymptote.

Be careful if a factor cancels. For example,

$$\frac{(x-2)(x+1)}{x-2}=x+1, \qquad x\ne2.$$

This is the straight line $y=x+1$ with a missing point at $(2,3)$, not a graph with a vertical asymptote at $x=2$.

The graph of $\boldsymbol{y=\frac{1}{x^2}}$

For every $x\ne0$,

$$\frac{1}{x^2}>0.$$

Both branches therefore lie above the $x$-axis. The function is even because

$$f(-x)=\frac{1}{(-x)^2}=\frac{1}{x^2}=f(x),$$

so its graph is symmetric about the $y$-axis. Its asymptotes are still $x=0$ and $y=0$.

This contrasts with $y=1/x$, which is an odd function and has rotational symmetry about the origin.

Exponential graphs

For $a>0$ and $a\ne1$, the exponential function is

$$y=a^x.$$

Every such graph passes through $(0,1)$ because $a^0=1$. Its values are always positive, so its range is $y>0$, and $y=0$ is a horizontal asymptote.

• If $a>1$, the function is increasing.
• If $0<a<1$, the function is decreasing.

Worked example 3: compare growth and decay

For $y=2^x$:

$$2^{-2}=\frac14, \quad 2^{-1}=\frac12, \quad 2^0=1, \quad 2^1=2, \quad 2^2=4.$$

The graph increases and approaches $y=0$ as $x\to-\infty$.

For $y=(1/2)^x=2^{-x}$, the same values appear in reverse order. Its graph decreases and approaches $y=0$ as $x\to\infty$.

Neither graph meets the $x$-axis. Writing a very small positive value as zero on a calculator display does not create an algebraic root.

Logarithmic graphs

The logarithmic function

$$y=\log_a x$$

is the inverse of $y=a^x$. Therefore its graph is the reflection of $y=a^x$ in the line $y=x$.

Its defining relationship is

$$y=\log_a x \quad\Longleftrightarrow\quad a^y=x.$$

Important features are:

• domain $x>0$;
• range all real numbers;
• vertical asymptote $x=0$;
• $x$-intercept $(1,0)$ because $\log_a1=0$;
• the point $(a,1)$ because $\log_a a=1$.

There is no $y$-intercept because $x=0$ is outside the domain.

Worked example 4: connect inverse graphs

The graph of $y=3^x$ contains the points

$$(-1,\tfrac13),\quad (0,1),\quad (1,3).$$

Reflecting in $y=x$ swaps each point’s coordinates, so $y=\log_3x$ contains

$$(\tfrac13,-1),\quad (1,0),\quad (3,1).$$

This coordinate swap is a useful check whenever two functions are inverses.

Modulus graphs

The modulus $|u|$ is the non-negative size of $u$:

$$|u|= \begin{cases} u, & u\ge0,\\ -u, & u<0. \end{cases}$$

Two superficially similar graph operations have different effects.

Outside the function: $\boldsymbol{y=|f(x)|}$

Any part of $y=f(x)$ below the $x$-axis is reflected above the $x$-axis. Parts already on or above the axis remain unchanged. The resulting range is non-negative.

Inside the function: $\boldsymbol{y=f(|x|)}$

Keep the part of $y=f(x)$ for $x\ge0$, then reflect that half in the $y$-axis. The original left half is discarded. The result is always an even function.

Worked example 5: distinguish the two modulus graphs

Let

$$f(x)=x-2.$$

Then

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

has a V-shaped graph with vertex $(2,0)$.

In contrast,

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

has vertex $(0,-2)$ and roots $x=-2$ and $x=2$.

The notation tells you which coordinates are altered. An outside modulus changes negative output values, while an inside modulus changes the input before the function acts.

Symmetry tests

Symmetry can reduce the work needed to sketch a graph.

Even functions

If

$$f(-x)=f(x),$$

then $f$ is even and its graph is symmetric about the $y$-axis. Examples include $x^2$, $x^4+3x^2$ and $\cos x$.

Odd functions

If

$$f(-x)=-f(x),$$

then $f$ is odd and its graph has rotational symmetry of $180^\circ$ about the origin. Examples include $x^3$, $x^5-2x$ and $\sin x$.

Worked example 6: test symmetry algebraically

Let

$$f(x)=x^4-3x^2+2.$$

Then

$$f(-x)=(-x)^4-3(-x)^2+2=x^4-3x^2+2=f(x).$$

The function is even.

Now let

$$g(x)=x^3+2x.$$

Then

$$g(-x)=(-x)^3+2(-x)=-x^3-2x=-g(x).$$

The function is odd.

A function can be neither even nor odd. Do not decide from one or two plotted points: apply the algebraic test to the whole expression.

Using graphs to solve equations

The solutions of

$$f(x)=g(x)$$

are the $x$-coordinates where the graphs $y=f(x)$ and $y=g(x)$ intersect.

Similarly, the solutions of $f(x)=k$ are where $y=f(x)$ meets the horizontal line $y=k$.

Worked example 7: count solutions without solving exactly

How many real solutions does

$$x^2=2^x$$

have?

Sketch $y=x^2$ and $y=2^x$ on the same axes. Some intersections can be identified exactly:

$$x=2: \quad x^2=4=2^2,$$ $$x=4: \quad x^2=16=2^4.$$

There is also one negative intersection because $x^2$ grows large as $x\to-\infty$, while $2^x\to0$, but at $x=0$ we have $x^2=0<1=2^x$. Therefore there are three real solutions in total.

The graph gives the number and approximate location of solutions even when an equation cannot be rearranged into a familiar exact form.

Common misconceptions

”A root and a $y$-intercept are the same”

A root has $y=0$ and lies on the $x$-axis. The $y$-intercept has $x=0$. They coincide only at the origin.

”Every graph crosses an asymptote”

Some graphs cross horizontal or oblique asymptotes and some do not. A vertical asymptote describes unbounded behaviour near a forbidden input. Check the function rather than relying on a slogan.

”A repeated root always crosses the axis”

Odd multiplicity roots cross. Even multiplicity roots touch and turn. The parity of the multiplicity is the deciding feature.

”A sketch should be made by plotting many decimal points”

A table of values can help, but it can miss roots, turning points and behaviour near asymptotes. Find exact structural features first, then use selected points only to confirm the shape.

”The graph ends at the edge of the axes”

The edge of a drawing is not part of the function. Use arrows or clear end behaviour to show that the curve continues.

Mixed practice

Question 1

For

$$f(x)=(x-4)^2(x+1),$$

state the intercepts, describe the behaviour at each root, and give the end behaviour.

Question 2

Find the asymptotes and intercepts of

$$y=\frac{3x-2}{x+1}.$$

Question 3

The point $(2,5)$ lies on $y=f(x)$. State a corresponding point on each graph:

1. $y=|f(x)|$;
2. $y=f(|x|)$;
3. $y=f^{-1}(x)$, assuming the inverse exists.

Question 4

Determine whether

$$h(x)=x^5-4x^3+x$$

is even, odd or neither.

Answers

1. The roots are $x=4$ and $x=-1$. The graph touches and turns at $(4,0)$ because $x=4$ is a repeated root, and crosses at $(-1,0)$. The $y$-intercept is

$$f(0)=(-4)^2(1)=16.$$

The leading term is $x^3$, so $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to\infty$ as $x\to\infty$.

2. Rearrange by division:

$$\frac{3x-2}{x+1}=3-\frac{5}{x+1}.$$

The asymptotes are $x=-1$ and $y=3$. The $x$-intercept is $(2/3,0)$ and the $y$-intercept is $(0,-2)$.

3. The point $(2,5)$ remains $(2,5)$ on $y=|f(x)|$ because its output is already positive. It remains $(2,5)$ on $y=f(|x|)$ and is mirrored to $(-2,5)$. On the inverse graph, its corresponding point is $(5,2)$.

4.

$$h(-x)=(-x)^5-4(-x)^3+(-x)=-x^5+4x^3-x=-h(x),$$

so $h$ is odd.

Final checklist

When you sketch a function graph, check that you have:

• used the correct domain;
• labelled exact intercepts where possible;
• distinguished crossing roots from touching roots;
• drawn and labelled any asymptotes;
• shown the correct end behaviour;
• respected symmetry;
• drawn a smooth curve where the function is smooth;
• included enough information for the sketch to be unambiguous.

What to learn next

• Learn how equations alter a graph in graph transformations.
• Connect a function with its reflection in $y=x$ in composite and inverse functions.
• Use derivatives to locate exact turning points in stationary points and curve sketching.
• Apply graph shapes to real situations in functions in modelling.