From 7e58f46f46f9e945a5ecb60534f166f8fad2082a Mon Sep 17 00:00:00 2001
From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Fri, 22 Jun 2018 10:42:36 +1000
Subject: [PATCH 1/1] start spec graphing techniques

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+# Graphing techniques
+
+## Reciprocal continuous functions
+
+If $y=f(x)$, the reciprocal function is:
+
+$$y={1 \over f(x)}$$
+
+As $\quad f(x) \rightarrow \pm \infty,\quad {1 \over f(x)} \rightarrow 0^\pm$ (vert asymptote at $f(x)=0$)
+
+As $\quad x \rightarrow  \pm \infty,\quad {-1 \over x}$
+
+- reciprocal functions are always on the same side of $x=0$
+- if $y=f(x)$ has a local max|min at $x=1$, then $y={1 \over f(x)}$ has a local max|min at $x=a$
+- point of inflection at $P(1,1)$
+
+## Locus of points
+
+- set of points that satisfy a given condition
+- path traced by a point that moves according to a condition
+
+### Circular loci
+
+$$(x-a)^2 + (y-b)^2 = r^2$$
+
+point $P(x,y)$ has a constant distance $r$ from point $C(a,b)$ (centre)
+
+### Linear loci
+
+$$QP=RP$$
+$$\sqrt{(x_Q-q_P)^2+(y_Q-y_P)^2} = \sqrt{(x_R-x_P)^2+(y_R-y_P)^2}$$
+
+points $Q$ and $R$ are fixed and have a perpendicular bisector $QR$. Therefore, any point on line $y=mx+c$ is equidistant from $QP$ and $RP$.
+
+Since the bisector of the line joining points $Q$ and $R$ is perpendicular to $QR$:
+
+$$m(QR) \times m(RP) = -1$$
+
+### Parabolic loci
+
+$$PD=PF$$
+$$|y-z|=\sqrt{(x-x_F)^2+(y-y_F)^2}$$
+$$(y-z)^2=(x-x_F)^2+(y-y_F)^2$$
+
+Distance of point $P(x,y)$ from fixed point $F(a,b)$ is equal to the distance of $P$ from $y=z \perp$.
+
+Fixed point $F$ is the **focus** (halfway between $y=z$ and $y=y_P$)
+
+Fixed line $x=z$ is the **directrix**
+
+### Elliptical loci
+
+$$F_1 P + F_2 P =k$$
+
+**Two** foci at $F_1$ and $F_2$
-- 
2.43.2