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.

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- Aug 8th 2011, 12:47 PMbenny92000Inverse 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 PMSironRe: 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 PMBERMES39Re: 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 PMBERMES39Re: Inverse functions and symmetry
Hey Siron we must have similar minds!

- Aug 8th 2011, 01:17 PMSironRe: Inverse functions and symmetry
- Aug 9th 2011, 07:37 AMbenny92000Re: Inverse functions and symmetry
So the two equations I are NOT inverses?

- Aug 9th 2011, 07:42 AMSironRe: 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?