Graphs of common functions: shapes and key features

A graph shows every pair of values that satisfies an equation. Recognising its overall shape is useful, but a mathematically meaningful sketch must also show its key features, such as intercepts, roots, turning points and asymptotes.

This lesson secures the graph knowledge needed before A level work on functions, transformations, coordinate geometry and calculus.

Before you begin

You should be able to:

• substitute numbers into algebraic expressions
• solve linear and quadratic equations
• expand and factorise quadratic expressions
• use positive, zero and negative indices
• plot coordinates and choose a sensible scale

Review algebraic fluency, quadratic equations or straight line graphs if needed.

What does the graph of a function mean?

For an equation

$$y=f(x),$$

each permitted input $x$ produces an output $f(x)$. The graph is the set of points

$$(x,f(x)).$$

For example, if $f(x)=x^2-1$, then

$$f(-2)=3,qquad f(0)=-1,qquad f(3)=8.$$

So $(-2,3)$, $(0,-1)$ and $(3,8)$ lie on the graph.

The graph is not just the few points in a value table. Unless the domain is discrete, it contains the points for all allowed values of $x$ between them.

Essential graph language

Feature	How to find it	Meaning
$y$-intercept	Set $x=0$	Where the graph meets the $y$-axis
$x$-intercept or root	Set $y=0$, so solve $f(x)=0$	Where the graph meets the $x$-axis
turning point	Identify where the curve changes from decreasing to increasing, or conversely	A local minimum or maximum
asymptote	Identify a line that the curve approaches	Describes behaviour near a forbidden value or far from the origin
positive values	Find where $f(x)>0$	Graph is above the $x$-axis
negative values	Find where $f(x)<0$	Graph is below the $x$-axis

Do not confuse an $x$-intercept $(a,0)$ with a $y$-intercept $(0,b)$. Coordinates must always be written in the order $(x,y)$.

Linear functions

A linear function has the form

$$y=mx+c.$$

Its graph is a straight line with gradient $m$ and $y$-intercept $(0,c)$. If $m>0$, the line rises from left to right. If $m<0$, it falls. If $m=0$, it is horizontal.

Worked example 1: find both intercepts

Find the intercepts of

$$y=3x-6.$$

For the $y$-intercept, set $x=0$:

$$y=3(0)-6=-6.$$

Hence the $y$-intercept is $(0,-6)$.

For the $x$-intercept, set $y=0$:

$$0=3x-6,qquad 3x=6,qquad x=2.$$

Hence the $x$-intercept is $(2,0)$. Plotting these two points is enough to determine the line.

See straight line graphs for gradients, parallel lines and equations of lines.

Quadratic functions

A quadratic function has the form

$$y=ax^2+bx+c,qquad a\ne0.$$

Its graph is a parabola.

• If $a>0$, the parabola opens upwards and has a minimum point.
• If $a<0$, it opens downwards and has a maximum point.
• Its axis of symmetry is the vertical line

$$x=-\frac{b}{2a}.$$

• It can have two, one or no $x$-intercepts.

The constant term gives the $y$-intercept immediately because

$$y=a(0)^2+b(0)+c=c.$$

Worked example 2: sketch from factorised form

Sketch the key features of

$$y=x^2-4x-5=(x-5)(x+1).$$

1. Find the roots.

$$(x-5)(x+1)=0$$

gives $x=5$ or $x=-1$. The graph crosses the axis at $(5,0)$ and $(-1,0)$.

2. Find the $y$-intercept.

Set $x=0$:

$$y=-5.$$

So the graph passes through $(0,-5)$.

3. Find the axis of symmetry.

The roots are equally spaced about the axis, so

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

4. Find the turning point.

Substitute $x=2$:

$$y=2^2-4(2)-5=-9.$$

The turning point is the minimum $(2,-9)$.

5. Complete the sketch.

The coefficient of $x^2$ is positive, so draw a smooth upward-opening curve through the three intercepts, symmetric about $x=2$, with minimum $(2,-9)$.

Worked example 3: a repeated root

Consider

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

Because $(x-3)^2=0$ only when $x=3$, there is one repeated root. The graph touches the $x$-axis at $(3,0)$ rather than crossing it. The negative sign makes the parabola open downwards, so $(3,0)$ is its maximum.

At $x=0$,

$$y=-(0-3)^2=-9,$$

so the $y$-intercept is $(0,-9)$.

Misconception: every quadratic crosses the axis twice

The graph of $y=x^2+4$ has minimum $(0,4)$, entirely above the $x$-axis. The equation

$$x^2+4=0$$

has no real solutions, so the graph has no $x$-intercepts.

Cubic functions

A cubic function has highest power $x^3$. The basic graph

$$y=x^3$$

passes through the origin, is increasing everywhere, and has rotational symmetry about the origin. Unlike a parabola, its ends point in opposite vertical directions.

A general cubic can have:

• one, two or three distinct $x$-intercepts
• no turning points or two turning points
• a repeated root where it touches the axis

Worked example 4: read a cubic from its factors

Describe the intercepts of

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

The roots are

$$x=-2\quad\text{and}\quad x=1.$$

At $x=-2$, the factor $x+2$ changes sign, so the graph crosses the axis at $(-2,0)$.

At $x=1$, the squared factor $(x-1)^2$ is non-negative on both sides. The graph touches the axis at $(1,0)$ and turns.

The $y$-intercept is

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

so it passes through $(0,2)$.

The leading term is $x\cdot x^2=x^3$. Therefore the graph goes down to the left and up to the right.

Crossing or touching?

For a factor $(x-a)^n$:

• an odd power usually makes the graph cross at $x=a$
• an even power makes the graph touch at $x=a$

At this stage, use this rule for factorised polynomial graphs. The fuller idea is called the multiplicity of a root.

Reciprocal functions

The reciprocal function

$$y=\frac1x$$

is undefined at $x=0$. Its graph has two separate branches, one in quadrant I and one in quadrant III.

The coordinate axes are asymptotes:

$$x=0\quad\text{and}\quad y=0.$$

As $x$ gets close to $0$, the magnitude of $1/x$ becomes very large. As $|x|$ gets very large, $1/x$ gets close to $0$.

$x$	$-10$	$-1$	$-0.1$	$0.1$	$1$	$10$
$1/x$	$-0.1$	$-1$	$-10$	$10$	$1$	$0.1$

The graph never contains a point with $x=0$, and $1/x$ is never zero. Therefore it has neither a $y$-intercept nor an $x$-intercept.

Worked example 5: the effect of a negative numerator

For

$$y=-\frac4x,$$

some useful points are

$$(-4,1),\quad (-2,2),\quad (2,-2),\quad (4,-1).$$

The branches lie in quadrants II and IV. The asymptotes remain $x=0$ and $y=0$.

Misconception: an asymptote is part of the graph

An asymptote guides the shape, but it is not part of $y=1/x$. Do not join the two branches through the origin. That would incorrectly assign a value when $x=0$.

Exponential functions

An exponential function has the variable in the exponent. For example,

$$y=2^x.$$

Since $2^x>0$ for every real $x$, its graph is always above the $x$-axis. Also,

$$2^0=1,$$

so it passes through $(0,1)$.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$2^x$	$\frac14$	$\frac12$	$1$	$2$	$4$	$8$

The line $y=0$ is a horizontal asymptote. The graph approaches it for increasingly negative $x$, but never reaches it.

If $0<a<1$, then $y=a^x$ represents exponential decay. For example,

$$\left(\frac12\right)^x=2^{-x}$$

decreases from left to right, still passes through $(0,1)$, and remains positive.

Worked example 6: distinguish exponential and quadratic growth

Compare $y=x^2$ and $y=2^x$ at several positive integer values.

$x$	$2$	$4$	$8$	$10$
$x^2$	$4$	$16$	$64$	$100$
$2^x$	$4$	$16$	$256$	$1024$

The graphs meet at $x=2$ and $x=4$, but that does not make them the same function. Eventually exponential growth outpaces polynomial growth because repeated multiplication is stronger than raising $x$ to a fixed power.

Circle graphs

The equation

$$x^2+y^2=r^2$$

describes a circle with centre $(0,0)$ and radius $r$.

This follows from Pythagoras’ theorem. Every point $(x,y)$ on the circle is distance $r$ from the origin, so

$$\sqrt{x^2+y^2}=r,$$

and squaring gives $x^2+y^2=r^2$.

Worked example 7: find intercepts and test a point

For

$$x^2+y^2=25,$$

the radius is $5$.

On the $x$-axis, $y=0$, so

$$x^2=25\implies x=\pm5.$$

The $x$-intercepts are $(-5,0)$ and $(5,0)$. Similarly, the $y$-intercepts are $(0,-5)$ and $(0,5)$.

To test $(3,4)$, substitute:

$$3^2+4^2=9+16=25.$$

Therefore $(3,4)$ lies on the circle.

Important distinction: not every graph is a function

The circle is a graph of an equation, but $x^2+y^2=25$ does not define $y$ as a function of $x$ over the whole circle. When $x=0$, both $y=5$ and $y=-5$ are possible. One input would have two outputs.

The vertical line test detects this: if any vertical line meets a graph more than once, that graph does not represent $y$ as a function of $x$.

A reliable sketching method

When asked to sketch a graph:

1. Identify the family from its equation.
2. Find the $y$-intercept by setting $x=0$, if permitted.
3. Find the $x$-intercepts by setting $y=0$.
4. Find turning points, symmetry and asymptotes where relevant.
5. Plot any extra easy points needed to fix the shape.
6. Draw a smooth curve consistent with every feature.
7. Label important coordinates and asymptotes.

A sketch does not usually need an exact scale. It does need the correct shape and correctly positioned key features.

Worked example 8: identify a function from its features

A graph:

• is symmetric about the $y$-axis
• has a minimum at $(0,3)$
• has no roots
• rises on both sides

Which of these could be its equation?

$$A:\ y=x^2+3,qquad B:\ y=x^3+3,qquad C:\ y=3^x,qquad D:\ y=\frac3x.$$

$A$ is the only possibility. The graph of $y=x^2+3$ is an upward-opening parabola translated three units upwards. Its minimum is $(0,3)$, so it never reaches the $x$-axis.

Common misconceptions

A small calculator value is not necessarily zero

For $y=2^x$, a calculator may display a very small decimal for a large negative $x$. Algebraically, $2^x$ is always positive and never equals zero.

A table does not determine the shape by itself

Joining plotted points with straight segments gives an inaccurate sketch of a curve. Use a table to locate points, then use knowledge of the function family to draw a smooth curve.

The equation decides whether a point is on a graph

A point that looks close to a curve is not necessarily on it. Substitute its coordinates into the equation and check whether both sides are equal.

Roots are input values, not points

If $x=4$ is a root, the corresponding intercept is the coordinate $(4,0)$. Write the form requested by the question.

Self-check

1. Intercepts

Find both intercepts of $y=2x+8$.

Answer

At $x=0$, $y=8$, so the $y$-intercept is $(0,8)$. At $y=0$,

$$2x+8=0\implies x=-4,$$

so the $x$-intercept is $(-4,0)$.

2. Quadratic features

State the roots, axis of symmetry and turning point of

$$y=(x-1)(x-5).$$

Answer

The roots are $x=1$ and $x=5$. Their midpoint is $3$, so the axis of symmetry is $x=3$. At $x=3$,

$$y=(3-1)(3-5)=2(-2)=-4.$$

The turning point is the minimum $(3,-4)$.

3. Match the family

Name the likely graph family in each case.

1. Two branches, no intercepts, asymptotes $x=0$ and $y=0$.
2. A smooth U-shaped curve with one minimum.
3. A curve through $(0,1)$ that is always positive and grows increasingly steeply.

Answer

1. Reciprocal, such as $y=1/x$.
2. Quadratic with positive leading coefficient.
3. Exponential growth, such as $y=2^x$.

4. Cubic roots

For

$$y=(x+3)^2(x-2),$$

where does the graph cross the $x$-axis, and where does it touch?

Answer

It touches at $(-3,0)$ because $x=-3$ is a repeated root of even power. It crosses at $(2,0)$ because $x=2$ comes from a factor of odd power.

5. Circle equation

Does $(5,12)$ lie on $x^2+y^2=169$?

Answer

Yes, because

$$5^2+12^2=25+144=169.$$

The circle has centre $(0,0)$ and radius $13$.

6. Reasoning challenge

A quadratic has roots $-2$ and $6$, and passes through $(0,-12)$. Find its equation.

Answer

The roots give

$$y=a(x+2)(x-6).$$

Use $(0,-12)$:

$$-12=a(2)(-6)=-12a,$$

so $a=1$. Therefore

$$\boxed{y=(x+2)(x-6)=x^2-4x-12}.$$

Next steps

• Learn how equations shift, reflect and stretch these shapes in transformations of graphs.
• Make the input and output language precise in function notation.
• Develop A level sketching, polynomial behaviour and rational graphs in graphs of functions.
• Connect roots and intersections with algebra in simultaneous equations and quadratic equations.