# Inverse functions and symmetry

• Aug 8th 2011, 12:47 PM
benny92000
Inverse functions and symmetry
I have the line y=ax + b
I am told that y=-ax + b is an inverse of the first line. Why is this so? I know that inverses are symmetrical about the line y=x. I graphed y=ax+3 and y=-ax+3, and they do not seem symmetrical about the line y=x to me.
• Aug 8th 2011, 12:58 PM
Siron
Re: Inverse functions and symmetry
Because the statement is not true, you can determine the inverse of $\displaystyle y=ax+b$ by changing $\displaystyle y$ and $\displaystyle x$ and solve back to y=.., like this:
$\displaystyle y=ax+b$
Calculating the inverse:
$\displaystyle x=ay+b \Leftrightarrow y=\frac{x-b}{a}$

For example:
$\displaystyle y=3x+2$
The inverse function has to be
$\displaystyle y=\frac{x-2}{3}$

Now look at the graphs of this functions, they're symmetrical towards the first bissectrice (y=x).
• Aug 8th 2011, 01:02 PM
BERMES39
Re: Inverse functions and symmetry
To find the inverse function of y= ax+b, replace x for y and y for x, then solve for y:
x=ay+b inverse: y= (x-b)/a
for instance: y = 3x +2 The inverse: y = (x-2)/3. Try the graph, they are symmetrical to the line y=x.
• Aug 8th 2011, 01:06 PM
BERMES39
Re: Inverse functions and symmetry
Hey Siron we must have similar minds!
• Aug 8th 2011, 01:17 PM
Siron
Re: Inverse functions and symmetry
Quote:

Originally Posted by BERMES39
Hey Siron we must have similar minds!

Yes, I was thinking the same! :)
• Aug 9th 2011, 07:37 AM
benny92000
Re: Inverse functions and symmetry
So the two equations I are NOT inverses?
• Aug 9th 2011, 07:42 AM
Siron
Re: Inverse functions and symmetry
$\displaystyle y=-ax+b$ is indeed not the inverse function of $\displaystyle y=ax+b$

Like you said, if you graph these lines then they're not symmetrical to the first bissectrice ($\displaystyle y=x$).

Have you graphed the example I and BERMES39 have given you?