AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsPolar Coordinates & Polar FunctionsGraphs of Polar Functions

Graphs of Polar Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 20 April 2026

Graphs of polar functions

How is a polar function defined?

A polar function is a function of the form $r = f (θ)$
- The input is an angle $θ$ (usually in radians)
- The output is a value $r$ that represents a signed radius
  - Remember that 'signed radius' means $r$ can be positive or negative
The graph of a polar function consists of all the points whose polar coordinates $(r, θ)$ satisfy the equation $r = f (θ)$
- For each input angle $θ$ in the domain, the point with polar coordinates $(f (θ), θ)$ is plotted
- As $θ$ varies over the domain, these points trace out a curve in the polar coordinate system
E.g. consider the following table of values for the polar function $r = 2 + 2 \cos θ$
- Note that when $θ = 2 π$ , the values go 'back to the beginning'
  - i.e. because $0$ and $2 π$ represent the same angle in a polar coordinate system

$θ$	$r = 2 + 2 \cos θ$
$0$	$4$
$\frac{π}{6}$	$2 + \sqrt{3} (\approx 3.732)$
$\frac{π}{3}$	$3$
$\frac{π}{2}$	$2$
$\frac{2 π}{3}$	$1$
$\frac{5 π}{6}$	$2 - \sqrt{3} (\approx 0.268)$
$π$	$0$
$\frac{7 π}{6}$	$2 - \sqrt{3} (\approx 0.268)$
$\frac{4 π}{3}$	$1$
$\frac{3 π}{2}$	$2$
$\frac{5 π}{3}$	$3$
$\frac{11 π}{6}$	$2 + \sqrt{3} (\approx 3.732)$
$2 π$	$4$

Plotting those points on a polar grid looks like this:

Polar graph with concentric circles and radial lines marked 1 to 5. Red dots are plotted at various coordinates across the grid. — **The points from the table plotted on a polar coordinates grid**

And connecting them with a smooth curve gives the graph of the polar function

Polar graph with a red cardioid curve. Concentric circles and radial lines form a grid. The polar axis is labelled, ranging from zero to five. — **Complete graph of r=2+2cosθ**

How do I interpret output values on a polar graph?

On the graph of $r = f (θ)$ , the behavior is different from what you are used to from a regular rectangular (Cartesian) coordinate system
Changes in the input $θ$
- correspond to changes in angle measure from the positive $x$ -axis
  - Remember that the positive $x$ -axis is called the polar axis in the polar coordinate system
Changes in the output $r$
- correspond to changes in signed distance from the origin
  - along the terminal ray of that angle $θ$
Because $r$ can be negative, the output values of a polar function are signed radius values, not just distances
- When $r > 0$ , the point lies on the terminal ray of the angle $θ$
  - at distance $r$ from the origin
- When $r < 0$ , the point lies on the opposite ray (direction reversed)
  - at distance $| r |$ from the origin
- When $r = 0$ , the point is at the origin
This means that as $θ$ changes
- the point on the graph can cross through the origin whenever $f (θ) = 0$
- and the graph can appear in a quadrant that is "opposite" to the direction of the terminal ray if $f (θ)$ becomes negative
Compare the following two graphs of polar functions
- In the first case, $r = 3 + 2 \cos θ$ is positive for all values of $θ$
  - So all the points occur on their terminal rays, in the same quadrants as their terminal rays
- In the second case, $r = 2 + 3 \cos θ$ becomes negative when $\cos θ < - \frac{2}{3}$ (or in the approximate range $2.301 < θ < 3.983$ , in the second and third quadrants)
  - For points with those $θ$ values, the points occur on the opposite rays to their terminal rays (and in the opposite quadrants to their terminal rays)
  - This is what creates the 'inner loop' on the graph

A polar graph featuring a grid of concentric circles, radial lines, and a red closed loop resembling a cardioid (sideways heart shape), with polar axes labelled from 1 to 5. — **Graph of 3+2cosθ**

A polar graph featuring a grid of concentric circles, radial lines, and a red outer loop resembling a cardioid (sideways heart shape) and a red inner loop, with polar axes labelled from 1 to 5. — **Graph of 2+3cosθ**

Examiner Tips and Tricks

On exam questions, keep an eye out for graphs showing curves with 'inner loops' like in the second graph above.

This is a clear sign that the function being graphed, $r = f (θ)$ , outputs negative values of $r$ for some values of $θ$
In a multiple choice question this may allow you immediately to rule out certain of the answer options

It is also possible for the output of a polar function to change while the distance of the point from the origin stays the same
- E.g. if $f (θ)$ changes from $- 2$ to $2$
  - the distance from the origin is $2$ in both cases
  - even though the output has changed
- This is similar to how, in the rectangular coordinate system
  - a change in $y$ -value from $- 2$ to $2$
  - does not change the distance from the $x$ -axis

How can the domain of a polar function be restricted?

A polar function is often given with a specified domain
- e.g. $0 \leq θ \leq 2 π$
  - For polar functions that appear in your exam, the full graph of the function will generally be traced out over the domain $0 \leq θ \leq 2 π$
If the domain is restricted to a smaller interval
- only the portion of the graph corresponding to angles in that interval remains
To identify what portion remains
- Find the endpoints of the restricted domain
  - These give the starting and ending points of the curve
- Trace out the curve between those endpoints
  - following the points $(f (θ), θ)$
  - as $θ$ increases across the interval
Pay special attention to angles
- where $f (θ) = 0$
  - the curve passes through the origin at such points
- and to where $f (θ)$ changes sign
  - the curve may jump to the opposite side of the origin at such points
The remaining portion may consist of more than one visible piece
- if the curve passes through the origin within the restricted interval

Examiner Tips and Tricks

When sketching or identifying a polar graph, always make a small table of values first.

Pick several angles in the domain (e.g. $0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, π, \dots$ )
and compute $r$ for each

Be especially careful with values of $θ$ where $f (θ)$ is negative

The resulting point appears on the opposite side of the origin from the terminal ray, not on the terminal ray itself
Forgetting this is one of the most common errors in identifying polar graphs

Worked Example

A polar function is defined by $f (θ) = 2 \sin θ + 1$ . Which of the following is the graph of the polar function $r = f (θ)$ in the polar coordinate system for $0 \leq θ \leq 2 π$ ?

(A)

Polar graph with a spiral line starting at the centre and looping outward, crossing concentric circles and radial lines numbered 1 to 5.

(B)

Polar graph with concentric circles from 1 to 5, lines at regular angles, and a thick, uneven graph tracing between 1 and 3. Labelled polar axis.

(C)

Polar graph with concentric circular grid lines centred on the origin and a plotted curve resembling a heart shape with an internal loop.

(D)

Polar graph with concentric circles centred at the origin and radial lines, displaying a thickly outlined curve, exhibiting one inner loop.

Answer:

Start examining the value of $r = f (θ)$ for different values of $θ$

When $θ = 0$

$f (0) = 2 \sin 0 + 1 = 1$

This rules out option (B), which doesn't go through $(r, θ) = (1, 0)$
Note as well that the inner loop on the option (A) graph does not make it immediately clear whether the point at $r = 1$ on the polar axis actually corresponds to $θ = 0$
- It could also come from $θ = π$ with a negative $r$ value

When $θ = \frac{π}{2}$

$f (\frac{π}{2}) = 2 \sin \frac{π}{2} + 1 = 3$

This rules out options (A) and (D), neither of which goes through $(r, θ) = (3, \frac{π}{2})$
Option (B) goes through that point, but that option was already ruled out in the previous step

That only leaves option (C), which must be the correct answer

If in doubt you could confirm this by computing $r = f (θ)$ for other values of $θ$
If you did that you would find that $r < 0$ for $\frac{7 π}{6} < θ < \frac{11 π}{6}$ , which is what creates the inner loop on the graph of (C)

The correct answer is (C)

Worked Example

Polar graph with concentric circles and radial lines, displaying a four-petaled rose curve centred at the origin.

The figure shows the graph of the polar function $r = f (θ)$ , where $f (θ) = 5 \cos (2 θ)$ , in the polar coordinate system for $0 \leq θ \leq 2 π$ . There are five points labeled $K, L, M, N$ , and $O$ , where $O$ is the origin. If the domain of $f$ is restricted to $π \leq θ \leq \frac{3 π}{2}$ , the portion of the given graph that remains consists of two pieces. One of those pieces is the portion of the graph in Quadrant III from $K$ to $O$ . Which of the following describes the other remaining piece?

(A) The portion of the graph in Quadrant I from $O$ to $L$

(B) The portion of the graph in Quadrant I from $O$ to $M$

(D) The portion of the graph in Quadrant IV from $O$ to $M$

Answer:

Note at the start that the terminal rays for $π \leq θ \leq \frac{3 π}{2}$ are all in Quadrant III

So points corresponding to those $θ$ values can only lie in Quadrant III (if values of $r = f (θ)$ are positive) or Quadrant I (if values of $r = f (θ)$ are negative)
This instantly rules out answer option (D)

Trace the values of $θ$ across the restricted domain $π \leq θ \leq \frac{3 π}{2}$

At $θ = π$

$r = 5 \cos (2 π) = 5$

The terminal ray for $θ = π$ points left from the origin (in the opposite direction to the polar axis)
and because $r$ is positive, the point lies on that ray at a distance of 5 from the origin
- This is point $K$
Note that $\cos (2 π) = \cos (0) = 1$

At $θ = \frac{5 π}{4}$

$r = 5 \cos (2 \cdot \frac{5 π}{4}) = 5 \cos (\frac{5 π}{2}) = 0$

This is point $O$ at the origin
Note that $\cos (\frac{π}{2}) = 0$ , and cosine has a period of $2 π$
- so $\cos (\frac{π}{2} + 2 π) = \cos (\frac{5 π}{2})$ is also equal to $0$

So going between $θ = π$ and $θ = \frac{5 π}{4}$ traces out "the portion of the graph in Quadrant III from $K$ to $O$ " that is mentioned in the question:

The graph from the question, with the portion of the curve in quadrant III between points K and O highlighted in red

For the portion of the graph between $θ = \frac{5 π}{4}$ and $θ = \frac{3 π}{2}$ there are only two possibilities:

If $r = f (θ)$ is positive then this portion of the graph will be in Quadrant III, tracing out the portion of the graph between $O$ and $N$
If $r = f (θ)$ is negative then this portion of the graph will be in Quadrant I (in the 'slice' of Quadrant I opposite the slice in Quadrant III between $θ = \frac{5 π}{4}$ and $θ = \frac{3 π}{2}$ ), tracing out the portion of the graph between $O$ and $L$

At $θ = \frac{3 π}{2}$

$r = 5 \cos (2 \cdot \frac{3 π}{2}) = 5 \cos (3 π) = - 5$

The terminal ray for $θ = \frac{3 π}{2}$ points down from the origin
However $r$ is negative, so the point lies on the ray opposite to that ray (i.e., up from the origin) at a distance of 5 from the origin
- This is point $L$
Note that $\cos (π) = - 1$ , and cosine has a period of $2 π$
- so $\cos (π + 2 π) = \cos (3 π)$ is also equal to $- 1$

This means that the second of the two possibilities mentioned above is the correct one

(A) The portion of the graph in Quadrant I from $O$ to $L$

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Graphs of Polar Functions (College Board AP® Precalculus): Revision Note

Graphs of polar functions

How is a polar function defined?

How do I interpret output values on a polar graph?

Examiner Tips and Tricks

How can the domain of a polar function be restricted?

Examiner Tips and Tricks

Worked Example

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis