Inverse functions and symmetry

• August 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.
• August 8th 2011, 12:58 PM
Siron
Re: Inverse functions and symmetry
Because the statement is not true, you can determine the inverse of $y=ax+b$ by changing $y$ and $x$ and solve back to y=.., like this:
$y=ax+b$
Calculating the inverse:
$x=ay+b \Leftrightarrow y=\frac{x-b}{a}$

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

Now look at the graphs of this functions, they're symmetrical towards the first bissectrice (y=x).
• August 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.
• August 8th 2011, 01:06 PM
BERMES39
Re: Inverse functions and symmetry
Hey Siron we must have similar minds!
• August 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! :)
• August 9th 2011, 07:37 AM
benny92000
Re: Inverse functions and symmetry
So the two equations I are NOT inverses?
• August 9th 2011, 07:42 AM
Siron
Re: Inverse functions and symmetry
$y=-ax+b$ is indeed not the inverse function of $y=ax+b$

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

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