Functions can be translated (shifted vertically or hoizontally without change of scale or shape), reflected (resulting in a curve that is the mirror image of another curve, where the 'mirror' may be an axis or a line should as y = x), and magnified (sretched or shrunk in one or more directions).

Sine and cosine functions have transformations similar to linear and quadratic functions.

Functions have transformations of the following types:

$f(x) + d$ | Vertical translation | Translation d units vertically upwards |

$f(x + c)$ | Horizontal translation | Translation c units horizontally to the left |

$-f(x)$ | Reflection across x-axis | Every value of y is made negative |

$f(-x)$ | Reflection across the y-axis | Every x-value is made negative |

$a⋅f(x)$ | Vertical stretch | Stretch by a factor of a in the vertical direction |

$f(bx)$ | Horizontal stretch | Stretch by a factor of 1/b in the horizontal direction |

The function $f(x)$ can be translated upwards (d), translated sideways (-c), stretched vertically (a), stetched horizontally ($1/b$), reflected in the $x$-axis (-f(x)) and reflected in the $y$-axis (f(-x)), by means of different constants in the equation: $y= af(bx - c) + d$

A function of the form $x↦{ax+b}/{cx+d}$, where $x≠d/c$, and $a$, $b$, $c$ and $d$ are constants, is called a simple rational function.

These functions have both a horizontal and vertical asymptote.

In general, the function $y = {ax + b}/{cx + d}$ can be rewritten to $y= A/{B(x-h)} + k$

where A is the vertical stretch factor, B is the reciprocal of the horizontal stretch factor, and $( \table h;k )$ is the translation in the $( \table x;y )$ directions.

Find the function $y=g(x)$ which results when transforming the function $x↦1/x$ by: a vertical stretch by factor $1/2$, a horizontal stretch by factor $3$, and a translation of $[\table 2;-3]$:

The vertical stretch of $1/2$ [general: ${f(x)}/2$] results in $1/{2x}$

The horizontal stretch of $3$ [general: ${f(x/3)}$] results in $3/{2x}$

The translation of of $[\table 2;-3]$ [general: ${f(x-2)-3}$] results in $3/{2(x-2)}-3$

$g(x) = 3/{2(x-2)}-3 = 3/{2x-4}-{3(2x-4)}/{(2x-4)} = {3-6x+12}/{(2x-4)} = {-3x+{15}/2}/{(x-2)}$

The asymptotes of $1/x$ are $x=0$ and $y=0$, so the asymptotes of the transformation are the translation parameters: $x-2=0$ ⇒ $x=2$, and $y=a/c={-3}/1=-3$

As the name suggests, invariant points do not move under a transformation. If there is no translation, a polynomial which passes through zero will still pass through zero irrespective of any stretching or inversions which occur.

The zeros of $f(x)$ become vertical asymptote values of $1/{f(x)}$

The vertical asymptote values of $f(x)$ become zeros of $1/{f(x)}$

Maximum values of $f(x)$ become minimum values of $1/{f(x)}$

Minimum values of $f(x)$ become maximum values of $1/{f(x)}$

When $f(x)>0$, $1/{f(x)}>0$

When $f(x)<0$, $1/{f(x)}<0$

When $f(x)→0$, $1/{f(x)}→±∞$

When $1/{f(x)}→0$, $f(x)→±∞$

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1687 - 1759

Nicolaus Bernoulli (I) was the first Nicolaus in the illustrious family dynasty of Bernoulli mathematicians in Basel, Switzerland, in the 17th and 18th centuries.

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