Transformation of functions

The vertex of $y=x^2$ sits at the origin. The picture wanted next puts that vertex at $(1,1)$, or makes the same arms twice as tall, or packs two bends of a sine in before the mark at $2\pi$, or turns a rising exponential into a falling one. Drawing each of those as a fresh assignment throws away the graph already in hand. A graph of a real function is only the set of pairs $(x,f(x))$, so the new picture has to be those pairs sent to new coordinates. Sent where?

The vertex leaves the origin

This case freezes the shape as $y=x^2$, and freezes the move as one step right and one step up.

The marked point starts at the vertex $(0,0)$ and lands on $(1,1)$. Every other point $(x,y)$ on $y=x^2$ lands on $(x+1,y+1)$, and the new graph is $y=(x-1{)}^2+1$.

The same sending with a free step $(a,b)$ is

$y=f(x-a)+b.$

The point now sitting at horizontal position $x$ is the one that used to sit at $x-a$, and its height is then lifted by $b$. The board's default $a=1$, $b=1$ is that vertex. Drag $a$ or $b$: the image is still the marked point plus $(a,b)$.

The vertex stays and the arms grow

$y=2x^2$ still has its vertex at the origin. The shift above left its vertex at $(1,1)$. Freeze the horizontal factor at $1$, and let only the vertical factor move. At $b=2$ the point $(t,t^2)$ goes to $(t,2t^2)$.

That graph is $y=2x^2$. At $b=1$ the two marks meet, and the extra height is gone.

With both factors free, and both required to be positive,

$y=bf\left(\frac{x}{a}\right),a>0,b>0.$

The old point $(t,f(t))$ goes to $(at,bf(t))$: factor $a$ along the horizontal axis, factor $b$ along the vertical axis.

The mark at $2\pi$ comes after two bends

Keep the vertical factor at $1$, and take the sine as the standard graph. Shorten the width to half, so $a=\frac{1}{2}$.

The image of a point keeps its height and takes half the input. Where $y=\sin x$ has finished one bend, the shortened graph has finished two, and the mark at $2\pi$ has not moved. The shortened graph is

$y=\sin \frac{x}{1/2}=\sin 2x.$

Set $a=1$ and the two curves lie on each other: the width is back to the sine you started with.

The exponential crosses the vertical axis

The stretch keeps both factors positive, so neither coordinate is replaced by its opposite. On $y=e^x$, send $(x,e^x)$ to $(-x,e^x)$.

Both curves pass through $(0,1)$. The image is $y=e^{-x}$, which is

$y=f(-x).$

The other sending keeps the input and changes the sign of the height. The point $(x,e^x)$ goes to $(x,-e^x)$.

That graph is $y=-f(x)$.

A point of a standard graph can be moved by adding $(a,b)$, rescaled by positive factors on the two axes, or sent to the opposite side of one axis. The same three sendings still apply when the standard graph is the line $y=x$.