# Thread: Inverse functions and symmetry

1. ## 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.

2. ## 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).

3. ## 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.

4. ## Re: Inverse functions and symmetry

Hey Siron we must have similar minds!

5. ## Re: Inverse functions and symmetry

Originally Posted by BERMES39
Hey Siron we must have similar minds!
Yes, I was thinking the same!

6. ## Re: Inverse functions and symmetry

So the two equations I are NOT inverses?

7. ## 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?