Reflecting the function $y=f(x)$ about $y=mx+b$ I want to reflect $y=f(x)$ about $y=mx+b$, and I'm struggling a bit to get one more piece of information. What I did so far was call my original point $(x_1,f(x_1))$ and call the point achieved by reflection $(x_2,g(x_2))$. I know that the line connecting these two points will be perpendicular to the line $y=mx+b$. Meaning that it will have a slope of $\frac{-1}{m}$:
$$\frac{g(x_2)-f(x_1)}{x_2-x_1}=\frac{-1}{m}$$
Now I'm trying to figure out how to get "$b$" into use, so I can solve $g(x_2)$ in terms of only $x_2$. Can someone help out or hint me to something, at this moment it's not that clear. 
 A: Allow me to get a bit more general.
Suppose we have a line $\ell$ in the plane characterized by $Ax+By+C=0,$ where $A,B,C$ are real and $A,B$ are not both $0$. Let's figure out how to reflect a point $\langle x_0,y_0\rangle$ about $\ell.$
First of all, we note/prove that for any $D\in\Bbb R,$ the line given by $$-Bx+Ay+D=0$$ is perpendicular to $\ell,$ and every line in the plane perpendicular to $\ell$ has such a form.
Second, letting $D=Bx_0-Ay_0,$ we have that $-Bx+Ay+D=0$ is the unique perpendicular of $\ell$ in the plane that passes through $\langle x_0,y_0\rangle.$
Third, we solve the system $$Ax+By+C=0\\-Bx+Ay+Bx_0-Ay_0=0$$ to obtain the unique $\langle x_1,y_1\rangle$ lying on $\ell$ and the given perpendicular.
Finally, we note/prove that $\langle x_2,y_2\rangle$ is the reflection of $\langle x_0,y_0\rangle$ about $\ell$ if and only if $\langle x_1,y_1\rangle$ is the midpoint of $\langle x_0,y_0\rangle$ and $\langle x_2,y_2\rangle,$ from which we can explicitly determine $\langle x_2,y_2\rangle.$
Since every line of the form $y=mx+b$ can be rewritten in the form $Ax+By+C=0,$ you should be able to proceed as such, with $y_0=f(x_0).$
It is worth noting that the reflection of a function in $x$ about a given line may not be a function in $x$! For example, consider the reflection of $f(x)=x^2$ about any line that isn't parallel to the $x$-axis.
