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+
# Graphing techniques
## Reciprocal continuous functions
As $\quad f(x) \rightarrow \pm \infty,\quad {1 \over f(x)} \rightarrow 0^\pm$ (vert asymptote at $f(x)=0$)
-As $\quad x \rightarrow \pm \infty,\quad {-1 \over x}$
+<!-- As $\quad x \rightarrow \pm \infty,\quad {-1 \over x}$ -->
+
+\includegraphics[width=0.25\textwidth]{./graphics/recip-parabola.png}
+\includegraphics[width=0.25\textwidth]{./graphics/recip-sin-cos.png}
- 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 $x=a$
- set of points that satisfy a given condition
- path traced by a point that moves according to a condition
+- graph on CAS - **conics**
### Circular loci
+point $P(x,y)$ has a constant distance $r$ from point $C(a,b)$ (centre)
+
+
+$$PC = r$$
+
$$(x-a)^2 + (y-b)^2 = r^2$$
-point $P(x,y)$ has a constant distance $r$ from point $C(a,b)$ (centre)
+
### Linear loci
-$$QP=RP$$
+$$QP = RP $$
$$\sqrt{(x_Q-q_P)^2+(y_Q-y_P)^2} = \sqrt{(x_R-x_P)^2+(y_R-y_P)^2}$$
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 $QR$:
+Since the bisector of the line joining points $Q$ and $R$ is perpendicular to $QR $:
-$$m(QR) \times m(RP) = -1$$
+$$m( QR ) \times m( RP ) = -1$$
### Parabolic loci
-$$PD=PF$$
+$$PD = PF $$
$$|y-z|=\sqrt{(x-x_F)^2+(y-y_F)^2}$$
$$(y-z)^2=(x-x_F)^2+(y-y_F)^2$$
### Elliptical loci
-$$F_1 P + F_2 P =k$$
+Point $P$ moves so that the sum of its distances from two fixed points $F_1$ and $F_2$ is a constant $k$.
+
+$${F_1 P} + F_2 P =k$$
**Two** foci at $F_1$ and $F_2$
+
+Cartesian equation for ellipses:
+$${(x-h)^2 \over a^2} + {(y-k)^2 \over b^2} = 1$$
+centered at $(h,k)$. Width is $2a$, height is $2b$.
+
+### Transformations
+$$(x,y) \rightarrow (x \prime, y \prime)$$
+
+where $x \prime$ and $y \prime$ are the transformation factors (dilation away from $x$-axis means coefficient of $y$ increases in $y \prime$, and vice versa).
+
+Transformed equation is the same as initial equation with each term divided by its dilation coefficients (must be in terms of $x\prime$ and $y\prime$).
+
+e.g.
+
+$x^2 + y^2 = 1$ is dilated $3$ from $x$, $5$ from $y$.
+Transformation rule is $(x\prime,y\prime) = (5x,3y)$
+$x={x\prime \over 5},\quad y={y\prime \over 3}$
+
+Equation $x^2 + y^2=1$ becomes
+
+$${(x\prime)^2 \over 25}+ {(y\prime)^2 \over 9}=1$$
+
+
+### Hyperbolic loci
+
+$$|(F_2P - F_1P )| = k$$
+
+Cartesian equation for hyperbolas ($a$ and $b$ are dilation factors):
+$${(x-h)^2 \over a^2} - {(y-k)^2 \over b^2} = 1$$
+
+Distance between vertices is $2a$
+Vertices given by $(h \pm a, k)$
+
+Asymptotes at $y=\pm {b \over a}(x-h)+k$
+To make hyperbola up/down rather than left/right, swap $x$ and $y$
+
+$y^2-x^2=1$ produces hyperbola shifted 90 $^\circ$ (top and bottom of asymptotes)
+
+## Parametric equations
+
+Parametric curve:
+
+$$x=f(t), \quad y=g(t)$$
+
+$t$ is the parameter
+
+To convert to cartesian, solve like simultaneous equations
+
+## Polar coordinates
+
+$$x = r\cos\theta, \quad y = r\sin\theta$$
+
+### Spirals
+$$r={\theta \over n\pi}$$
+- solve intercepts for multiples of $\pi \over 2$
+- or draw table of values for $r$ and $\theta$ for each $n\pi \over 2$
+
+### Circles
+$$r=a$$
+
+### Lines
+
+Horizontal: $r={n \over \sin \theta}$
+Vertical: $r={n \over \cos \theta}$
+
+### Cardioids
+
+$$r=a(n+ \cos\theta)$$
+
+### Roses
+
+$$r=\cos(k\theta)$$
+
+If $k$ is odd, half of the petals will overlap (hence there are $n$ petals)
+
+If $k$ is even, petals will not overlap (hence $2n$ petals)
+
+\includegraphics[width=0.5\textwidth]{./graphics/rose.png}
+
+
+### Solving polar graphs
+
+solve in terms of $r$
+
+e.g. $x=4$
+
+$r\cos\theta = 4$
+
+$r={4 \over \cos\theta}$
+
+
+---
+
+e.g. $y=x^2$
+
+$r\sin\theta = r^2 \cos^2 \theta$
+
+$\sin \theta = r \cos^2 \theta$
+
+$r = {\sin \theta \over \cos^2\theta} = \tan\theta \sec\theta$
+
+---
+
+e.g. $r=6\cos \theta\quad$ *(multiply by $r$)*
+
+$r^2=6r\cos\theta$
+
+$x^2+y^2=6x$
+
+complete the square