This page contains a bunch of graphs and some awesome graph drawing demos made by a friend of mine.

Basic Stuff

Let's say that $f$ is a function and we have a graph $y=f(x)$. It means that we can get a $y$ value by plugging the corresponding $x$ value to $f$. For example, at $x=1.5$, the $y$ value is $f(1.5)$.

There are a few common ways to stretch and move graphs:

Think about it like this: if we have $f(x-2)$ instead of $f(x)$, then $x$ needs to be bigger for getting the same $y$ value, and bigger $x$ means right. For example, $f(3) = f(5-2)$, so whatever was at $x=3$ moves to $x=5$.


The equation of any (non-vertical) line is $y=sx+c$, where $s$ and $c$ are constants. The $s$ is known as the slope and $c$ is the constant term. Try at least these things with the below demo program:

See Also

The derivative tutorial contains more information about slopes.


The "tip" of a parabola is known as the vertex. See the image at right.

The graph $y=x^2$ is a simple parabola, and the below demo displays that by default. As you can see, the vertex $x$ and $y$ coordinates are both 0. Parabolas of the form $y=(x-x_v)^2+y_v$ are the same thing shifted, and the $x_v$ and $y_v$ constants represent new $x$ and $y$ coordinates of the vertex.

Putting a constant $a$ to front like $y=a\cdot(x-x_v)^2+y_v$ makes the parabola stretch, and a negative $a$ means that it opens down. Note that $a \ne 0$ because $a=0$ would mean $y=y_v$, and that's a horizontal line at height $y_v$ instead of a parabola.

Handy thing: $(a-b)^2=a^2-2ab+b^2$

Proof using $(a-b)c=ac-bc$:

$$\begin{align}& \ (a-b)^2 \\ =&\ (a-b)(a-b) \\ =&\ a(a-b)-b(a-b) \\ =&\ (aa-ab)-(ba-bb) \\ =&\ aa-ab-ab+bb \\ =&\ aa-(ab+ab)+bb \\ =&\ a^2-2ab+b^2\end{align}$$

In this case we need to plug in $a=x$ and $b=x_v$.

The $y=a(x-x_v)^2+y_v$ form is nice because it's easy to see where the vertex is and it's easy to solve things like $y=0$, but let's see what happens if we expand that:

$$\begin{align}y &= a\cdot(x-x_v)^2+y_v \\ &= a\cdot(x^2-2x_vx+{x_v}^2)+y_v \\ &= ax^2-2ax_vx+a{x_v}^2+y_v\end{align}$$

Now if we set $b=-2ax_v$ and $c=a{x_v}^2+y_v$, then $b$ and $c$ are also constants because they don't depend on the value of $x$. We get this:

$$\begin{align}y = ax^2+bx+c\end{align}$$

This form comes up often when calculating stuff, and converting from this form back to $y=a(x-x_v)^2+y_v$ can be useful. That is known as completing the square. This exercise gives you everything you need to know for doing that:


Above we set $b=-2ax_v$ and $c=a{x_v}^2+y_v$. If we know $a$, $b$ and $c$, how can we figure out $x_v$ and $y_v$?

Check your $x_v$ answer with a derivative like $\frac{d}{dx}(ax^2+bx+c)=0$, and check your $y_v$ answer by plugging in $x=x_v$ to $y=ax^2+bx+c$.

Next, plug in your answers to this:

$$\begin{align}ax^2+bx+c = a(x-x_v)^2+y_v\end{align}$$

You should get the same square-completing formula as the one on the summary page.

Exponents and Logarithms

There's a detailed introduction to exponents and logarithms here. In short, $\log_b(x)$ is the inverse of $b^x$:

$$\begin{align}\log_b(b^x) &= x \\ b^{\log_b(x)} &= x\end{align}$$

Try at least these things with the below demos:

You're probably wondering why the base of a logarithm cannot be 1, 0 or negative. You'll see why if you draw the graph of $y=1^x$. If a logarithm with base 1 exists, then by definition, $y=1^x$ is same as $x=\log_1(y)$. But $x$ gets every possible value when $y=1$ and no values at all otherwise because $y=1^x$ is a horizontal line at height 1. A function like that doesn't really make much sense.

Sine Waves

This section assumes that you know how sine and cosine are defined in a unit circle. I have explained it in this chapter.

Here's a unit circle and a simple sine graph:

If you don't understand something about these things (e.g. what is $\sin(\theta)$, what is $\tau$, how do these graphs relate to each other, why does the blue graph look like a wave), please check those things from this chapter before you continue. The graph of cosine is a lot like the sine graph above, but we'll focus on the sine wave for now.

First, let's notice that the sine wave repeats itself with intervals of $\tau$:

This makes sense because if we rotate a full turn on the unit circle we get right back to where we started, and sine gets the same value. The repeating interval is known as the period $T$.

$$\begin{align}\sin(x+\tau) = \sin(x)\end{align}$$

If we graph $y=\sin(\tau x)$ instead of $y=\sin(x)$, we get a period of $1$ instead of $\tau$. Think about it like this: if $x$ changes by 1, then $\tau x$ changes by $\tau$ and $\sin(\tau x)$ repeats.

$$\begin{align}\sin(\tau(x+1)) = \sin(\tau x + \tau) = \sin(\tau x)\end{align}$$

In general, if we graph $y=\sin(\frac{\tau x}{T})$ we get a period $T$:

$$\begin{align}\sin\biggl(\frac{\tau(x+T)}{T}\biggr) &= \sin\biggl(\frac \tau T (x+T)\biggr) = \sin\biggl(\frac \tau T x + \frac{\tau}{\rcancel T}\rcancel T\biggr) \\ &= \sin\biggl(\frac{\tau x}{T} + \tau\biggr) = \sin\Bigl(\frac{\tau x}{T}\Bigr)\end{align}$$

This fun demo lets you play with $y=A\sin(\frac{\tau x}{T})$ graphs. The amplitude $A$ is just stretching the graph away from the $x$ axis.

The demo lets you make $A$ and $f$ negative, but as you can see, that just flips the graph around and it still looks very similar; this is why $A$ and $f$ are usually chosen to be positive.