Graphing

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 25 July 2012

Basic graphs

Lines

Parabolas

The Basic Circle

What is the equation of the basic circle?

Distances:

For the moment, call it $r$.

So

$\begin{array}{ccccc}\text{Area of the outer square}& =& \text{Area of the inner square}& +& \text{Area of 4 triangles}\\ {\left(p+q\right)}^2& =& r^2& +& 4·\frac{1}{2}pq\end{array}$

This is the heavily used Pythagorean Theorem.

So what is the equation of the basic circle?

$\begin{array}{ccc}\sqrt{x^2+y^2}=1& \text{is}& \text{all points}\left(x,y\right)\text{that are distance 1 from the origin.}\end{array}$

So $x^2+y^2=1$ is the equation of the basic circle.

The Basic Hyperbola $x^2-y^2=1$

$\begin{array}{cc}\multicolumn{2}{c}{\text{Graphing Notes}}\\ \text{(a)}& \text{If }y=0\text{ then }{x}^2=1\text{. So }x=\pm 1\text{.}\\ \text{(b)}& \text{If }x=0\text{ then }-y^2=1\text{ which is impossible.}\\ \text{(c)}& \text{The equation is }1-{\left(\frac{y}{x}\right)}^2={\left(\frac{1}{x}\right)}^2\text{(divide both sides by }{x}^2\text{).}\end{array}$

If $x$ gets very big ${\displaystyle \frac{1}{x}}$ gets closer and closer to 0 and the equation gets closer and closer to ${\displaystyle 1-{\left(\frac{y}{x}\right)}^2=0}$. This is the same as ${\displaystyle {\left(\frac{y}{x}\right)}^2=1}$, which is the same as ${\displaystyle \frac{y}{x}=\pm 1},\; i.e. y=\pm x$. So as $x$ gets very large the equation gets closer and closer to $y=x$ and $y=-x$.

As $x$ gets very negative the basic hyperbola gets closer and closer to $y=x$ and $y=-x$.

$\begin{array}{cc}\multicolumn{2}{c}{\text{Asymptotes}}\\ y=x\text{is an asymptote of the basic hyperbola as }x\to \infty \\ y=-x\text{is an asymptote of the basic hyperbola as }x\to +\infty \\ y=x\text{is an asymptote of the basic hyperbola as }x\to -\infty \\ y=-x\text{is an asymptote of the basic hyperbola as }x\to -\infty \end{array}$

An asymptote of a graph $y=f\left(x\right)$ as $x\to a$ is another graph $y=g\left(x\right)$ that the original graph $y=f\left(x\right)$ gets closer and closer to as $x$ gets closer and closer to $a$.

Example:    Graph  ${\displaystyle y=\frac{1}{x}}$

$\begin{array}{cc}\text{(a)}& \text{As }x\text{ gets large }\frac{1}{x}\text{ gets closer and closer to 0.}\\ \text{(b)}& \text{As }x\text{ gets closer to 0 (from the positive side) }\frac{1}{x}\text{ gets larger and larger.}\\ \text{(c)}& \text{As }x\text{ gets closer to 0 (from the negative side) }\frac{1}{x}\text{ gets more and more negative.}\\ \text{(d)}& \text{As }x\text{ gets more and more negative }\frac{1}{x}\text{ gets closer and closer to 0.}\\ \text{(e)}& \text{If }x=1\text{ then <math><mi>y</mi><mo>=</mo><mn>1</mn></math>}\\ \text{(f)}& \text{If }x=-1\text{ then <math><mi>y</mi><mo>=</mo><mo>-</mo><mn>1</mn></math>}\end{array}$

$\begin{array}{cc}\multicolumn{2}{c}{\text{Asymptotes}}\\ y=0\text{(The }x\text{ axis) is an asymptote to }y=\frac{1}{x}\text{ as }x\to \infty \\ y=0\text{(The }x\text{ axis) is an asymptote to }y=\frac{1}{x}\text{ as }x\to -\infty \\ x=0\text{(The }y\text{ axis) is an asymptote to }y=\frac{1}{x}\text{ as }x\to 0^{+}\\ x=0\text{(The }y\text{ axis) is an asymptote to }y=\frac{1}{x}\text{ as }x\to 0^{-}\end{array}$

Shifting

Lines

Parabolas

Circles

Hyperbolas

Scaling

Lines

Circles/Ellipses

Parabolas

Hyperbolas

Scaling and Shifting

Circles/Ellipses

Hyperbolas

Finding the shifts

$\begin{array}{cc}{\left(x-3\right)}^2=x^2-6x+9& \text{Note:}3=\frac{6}{2}\\ {\left(x-4\right)}^2=x^2-8x+16& 4=\frac{8}{2}\\ {\left(x-5\right)}^2=x^2-10x+25& 5=\frac{10}{2}\\ {\left(x-40\right)}^2=x^2-80x+1600\\ {\left(x-38\right)}^2=x^2-76x+{38}^2\\ {\left(x-\frac{b}{2}\right)}^2=x^2-bx+{\left(\frac{b}{2}\right)}^2\end{array}\begin{array}{c}\text{We can go backwards:}\\ x^2-6x+31=x^2-6x+9+31-9={\left(x-3\right)}^2+22\\ x^2-8x+15=x^2-8x+16+15-16={\left(x-4\right)}^2-1\\ x^2-10x-3=x^2-10x+25-3-25={\left(x-5\right)}^2-28\\ x^2-76x+18611=x^2-76x+{38}^2+18611-{38}^2={\left(x-38\right)}^2+17167\\ x^2-bx+c=x^2-bx+{\left(\frac{b}{2}\right)}^2+c-{\left(\frac{b}{2}\right)}^2={\left(x-\frac{b}{2}\right)}^2+c-{\left(\frac{b}{2}\right)}^2\\ \text{This procedure is called <strong>completing the square</strong>.}\\ \\ \text{Example:}\text{Graph}y=x^2-6x+2\\ y=x^2-6x+2=x^2-6x+9+2-9={\left(x-3\right)}^2-7\\ \text{So this is the same problem as graphing}\\ y+7={\left(x-3\right)}^2\\ \text{which is the basic parabola }y=x^2\text{ except adding 7 to }y\text{ shifts down by 7 and subtracting 3 from }x\text{ shifts right by 3.}\end{array}$

$\begin{array}{c}\text{Example:}\\ 9x^2-72x+y^2+4y+139=0\\ {\displaystyle x^2-8x+\frac{y^2+4y+139}{9}=0}\\ {\displaystyle x^2-8x+16+\frac{y^2+4y+4}{9}+\frac{139}{9}-16-\frac{4}{9}=0}\\ {\displaystyle {\left(x-4\right)}^2+\frac{{\left(y+2\right)}^2}{9}+\frac{139-144-4}{9}=0}\\ {\displaystyle {\left(x-4\right)}^2+{\left(\frac{y+2}{3}\right)}^2+\frac{-9}{9}=0}\\ {\displaystyle {\left(x-4\right)}^2+{\left(\frac{y+2}{3}\right)}^2=1}\\ \text{So this is the same as the basic circle }{x}^2+y^2=1\text{ except the center is shifted to }\left(4,-2\right)\text{ and the }y\text{ axis is stretched by 3.}\\ \\ \text{Example:}a{x}^2+bx+c{y}^2+dy+e=0\\ {\displaystyle \text{Is the same as}a\left(x^2+\frac{b}{a}x\right)+c\left(y^2+\frac{d}{c}y\right)=-e\text{.}}\\ {\displaystyle a\left(x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2\right)+c\left(y^2+\frac{d}{c}y+{\left(\frac{d}{2c}\right)}^2\right)=-e+a{\left(\frac{b}{2a}\right)}^2+c{\left(\frac{d}{2c}\right)}^2}\\ {\displaystyle a{\left(x+\frac{b}{2a}\right)}^2+c{\left(y+\frac{d}{2c}\right)}^2=-e+\frac{b^2}{4a}+\frac{d^2}{4c}}\\ {\displaystyle \frac{{\left(x+\frac{b}{2a}\right)}^2}{\frac{1}{a}}+\frac{{\left(y+\frac{d}{2c}\right)}^2}{\frac{1}{c}}=\frac{b^{2}c+d^{2}a-4ace}{4ac}}\\ {\displaystyle \frac{{\left(x+\frac{b}{2a}\right)}^2}{\left(\frac{b^{2}c+d^{2}a-4ace}{4a^{2}c}\right)}+\frac{{\left(y+\frac{d}{2c}\right)}^2}{\left(\frac{b^{2}c+d^{2}a-4ace}{4a{c}^2}\right)}=1}\\ {\displaystyle {\left(\frac{x+\frac{b}{2a}}{\sqrt{\frac{b^{2}c+d^{2}a-4ace}{4a^{2}c}}}\right)}^2+{\left(\frac{y+\frac{d}{2c}}{\sqrt{\frac{b^{2}c+d^{2}a-4ace}{4a{c}^2}}}\right)}^2=1}\\ \text{which is the basic circle}{x}^2+y^2=1\text{except}\\ {\displaystyle \text{the center is shifted to}\left(\frac{-b}{2a},\frac{-d}{2c}\right)}\\ {\displaystyle \text{the }x\text{ axis is stretched by}\sqrt{\frac{b^{2}c+d^{2}a-4ace}{4a^{2}c}}}\\ {\displaystyle \text{the }y\text{ axis is stretched by}\sqrt{\frac{b^{2}c+d^{2}a-4ace}{4a{c}^2}}}\\ \\ \text{It might help to compare this with }\text{exactly}\text{ the same example when}a=2\text{, }b=12\text{, }c=3\text{, }d=30\text{, }e=-3\text{.}\\ \text{So}2x^2+12x+3y^2+30y-3=0\text{,}\text{is the same as}\\ 2\left(x^2+6x\right)+3\left(y^2+10y\right)=0\\ 2\left(x^2+6x+9\right)+3\left(y^2+10y+25\right)=3+2·9+3·25\\ 2{\left(x+3\right)}^2+3{\left(y+5\right)}^2=3+18+75\\ {\displaystyle \frac{{\left(x+3\right)}^2}{\frac{1}{2}}+\frac{{\left(y+5\right)}^2}{\frac{1}{3}}=96}\\ {\displaystyle \frac{{\left(x+3\right)}^2}{\frac{96}{2}}+\frac{{\left(y+5\right)}^2}{\frac{96}{3}}=1}\\ {\displaystyle \frac{{\left(x+3\right)}^2}{\frac{96}{2}}+\frac{{\left(y+5\right)}^2}{\frac{96}{3}}=1}\\ {\displaystyle {\left(\frac{x+3}{\sqrt{\frac{96}{2}}}\right)}^2+{\left(\frac{y+5}{\sqrt{\frac{96}{3}}}\right)}^2=1}\\ {\displaystyle {\left(\frac{x+3}{\sqrt{48}}\right)}^2+{\left(\frac{y+5}{\sqrt{32}}\right)}^2=1}\\ \text{which is the basic circle}{x}^2+y^2=1\text{except}\\ {\displaystyle \text{the center is shifted to}\left(-3,-5\right)}\\ {\displaystyle \text{the }x\text{ axis is stretched by}\sqrt{48}\text{,}\text{and}}\\ {\displaystyle \text{the }y\text{ axis is stretched by}\sqrt{32}\text{.}}\end{array}$

Notes and References

These notes are from MATH 272 Lectures 1 and 2 given by Arun Ram on Jan 25 2000, Jan 27 2000.

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