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.
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.
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).
$\displaystyle y=-ax+b$ 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?