# VCE Specialist Mathematics/Units 3 and 4: Specialist Mathematics/Coordinate Geometry
## Preface

Formal Definition: Coordinate (algebraic) geometry is a branch of mathematics which combines techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry.
Translation: Understanding the math behind the various features that appear on a graph, allowing one to rapidly, and accurately, draw complex graphs (with points of interest).

## Properties of Graphs
### Asymptotes
#### Definition

Asymptotes are values which the graph approaches but does not touch. An asymptote is itself a graph and is categorized as follows:
* Vertical; A constant value (graph) on the horizontal axis (e.g. $x=1$).
* Horizontal; A constant value (graph) on the vertical axis (e.g. $y=1$).
* Oblique (i.e. not Vertical or Horizontal); A non-constant graph (e.g. $y=x^{2}$).

#### Comprehension

* Take a function $y=ax^{m}+{\frac {1}{bx^{n}}}\ +c$.
* As $x\to 0,{\frac {1}{bx^{n}}}\ \to \infty ,ax^{m}+c\to c$
* The numerator (a function or a number, as is shown here: $1$) is divided by an extremely small number. Hence making the fraction an extremely large number.
* To understand why this happens grab any number, and divide it by an extremely small number (e.g. $10^{-30}$)
* The value of ${\frac {1}{bx^{n}}}\ \to \infty$ overshadows the rest of the graph, namely the $ax^{m}+c\to c$.
* Hence making ${\frac {1}{bx^{n}}}\$ an oblique (non-constant) asymptote, as it is approached by, but never actually touched by the graph due to the addition of $ax^{m}+c$ to every y value. This is in addition to the limit provided when $x=1$ which is not included in this graph, but must be if it exists on others (e.g. after polynomial division).
* As $x\to \infty ,{\frac {1}{bx^{n}}}\ \to 0,ax^{m}+c\to \infty$
* The numerator (a function or a number, as is shown here:$1$) is divided by an extremely large number. Hence making the fraction an extremely small number.
* To understand why this happens grab any number, and divide it by an extremely large number (e.g. $10^{30}$)
* The value of $ax^{m}+c\to \infty$ overshadows the rest of the graph, namely the ${\frac {1}{bx^{n}}}\ \to 0$.
* Hence making $ax^{m}+c$ the oblique (non-constant) asymptote, as it is approached by, but never actually touched by the graph due to the addition of ${\frac {1}{bx^{n}}}\$ to every y value.

### Circles, Ellipses and Hyperbolas
#### Definition
##### Ellipses

General formula:
* ${\frac {(x-h)^{2}}{a^{2}}}+{\frac {(y-k)^{2}}{b^{2}}}=1$
General Notes:
* Point $(h,k)$ defines the ellipses center.
* Points $(\pm a+h,k)$ defines the ellipses domain, and horizontal endpoints - i.e. horizontal stretch.
* Points $(h,\pm b+k)$ defines the ellipses range, and vertical endpoints - i.e. vertical stretch.

##### Circles

General formula:
* $(x-h)^{2}+(y-k)^{2}=r^{2}$
General Notes:
* Point $(h,k)$ defines the circles center.
* Points $(\pm r+h,k)$ defines the circles domain - i.e. stretch.
* Points $(h,\pm r+k)$ defines the circles range - i.e. stretch.
* A circle is a subset of an ellipse, such that $a=b=r$.

##### Hyperbolas

General formulae:
* ${\frac {(x-h)^{2}}{a^{2}}}-{\frac {(y-k)^{2}}{b^{2}}}=1$
* ${\frac {(y-k)^{2}}{b^{2}}}-{\frac {(x-h)^{2}}{a^{2}}}=1$
General Notes:
* Point $(h,k)$ defines the hyperbolas center.
* Points $(\pm a+h,k)$ defines the hyperbolas domain, $[\pm a+h,\pm \infty )$.
* The switch in positions of the fractions containing x and y, indicate the type of hyperbola - i.e. vertical or horizontal. The hyperbola is horizontal in the first, and negative in the second of the General hyperbolic formulae above.
* Graphs $y=\pm (\pm a+h,k)$ defines the hyperbolas domain $[\pm a+h,\pm \infty )$.

#### Comprehension

Is unnecessary for these types of graphs, as the rules listed above are all that are needed. However the ability to recognize these graphs, in varying forms, is required and can be achieved via practice or can be found in Coordinate Geometry section of Common Math Hacks.

### Partial Fractions
#### Graphing Examples
##### General Steps

* Note the limits caused by a divide by 0.
* If possible (check the highest power of bottom and top), break up the complex function via polynomial division or partial fractions.
* If numerator's power is $<$ the denominators power, utilize partial fractions.
* Otherwise if the numerator's power is $\geq$ the denominators power, utilise polynomial division.
* Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
* Determine the asymptotes (Vertical, Horizontal, Oblique (i.e. a graph)).
* Determine other points of interest (Turning points (differentiation), Intercepts (let $y=0$, or $x=0$)).
* Draw the Graph using the above properties.

##### Partial Fractions

* Take a function: $y={\frac {3x+7}{x^{2}-2x-3}}\$
* Notice that if $x^{2}-2x-3=0$, hence $(x+1)(x-3)=0$, you get a divide by 0. Hence $x\neq -1$ or $x\neq 3$.
* Break up the function into partial fractions, and you arrive at $y={\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\$.
* Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
* Notice that as $x\to -1,{\frac {-1}{(x+1)}}\ \to -\infty ,{\frac {4}{(x-3)}}\ \to -1$. The fractional part, ${\frac {-1}{(x+1)}}\$, overshadows the rest of the equation, namely ${\frac {4}{(x-3)}}\$. Hence a horizontal asymptote occurs when $x=-1$.
* Notice that as $x\to 3,{\frac {4}{(x-3)}}\ \to \infty ,{\frac {-1}{(x+1)}}\ \to {\frac {-1}{4}}\$. The fractional part, ${\frac {4}{(x-3)}}\$, overshadows the rest of the equation, namely ${\frac {-1}{4}}\$. Hence a horizontal asymptote occurs when $x=3$.
* Notice that as $x\to \infty ,{\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\ \to 0,y\to 0$. Now the oblique (i.e. graph), $y=0$, overshadows the rest of the equation, namely ${\frac {-1}{(x+1)}}\ +{\frac {4}{(x-3)}}\$. Hence the oblique asymptote occurs when $y=0$.
* Determine points of interest:
** When $y=0,x={\frac {-7}{3}}\$, hence there are no x-intercepts (C is the complex field) in the real plane.
** When $x=0,y={\frac {-7}{3}}\$.
** When $y'=0,x=-5orx={\frac {1}{3}}\ ,y={\frac {-1}{4}}\ or{\frac {-9}{4}}\$
* Draw the graph.

##### Polynomial Division

* Take a function: $y=3x+7+{\frac {x^{3}+19x^{2}+116x+224}{x(x+4)(x+7)}}\$
* Notice that if $x(x+4)(x+7)=0$, you get a divide by 0. Hence $x\neq 0$ or $x\neq -4$ or $x\neq -7$
* Break up the function, and divide through, using polynomial division, and you arrive at $y=3x+7+{\frac {x+8}{x}}\$.
* Add the resultant graphs, through the addition of ordinates method, to quickly determine what the graph looks like.
* Notice that as $x\to 0,{\frac {x+8}{x}}\ \to \infty ,3x+7\to 7$. The fractional part, ${\frac {x+8}{x}}\$, overshadows the rest of the equation, namely $3x+7$. Hence the horizontal asymptote occurs when $x=0$.
* Notice that as $x\to \infty ,{\frac {x+8}{x}}\ \to 0,y\to 3x+7$. Now the oblique (i.e. graph), $3x+7$, overshadows the rest of the equation, namely ${\frac {x+8}{x}}\$. Hence the oblique asymptote occurs when $y=3x+7$.
* Determine points of interest:
** When $y=0,x\in \mathbb {C}$, hence there are no x-intercepts (C is the complex field) in the real plane.
** When $x=0,y\uparrow$, hence there are no y-intercepts ($\uparrow$ indicates that the previous statement is undefined.)
** When $y'=0,x=\pm {\frac {2{\sqrt {6}}}{3}}\ ,y=8\pm 4{\sqrt {6}}$
* Draw the graph.