Consider the graph of $y = e^x$
(a) Find the equation of the graph that results from reflecting about the line $y = 4$.
(b) Find the equation of the graph that results from reflecting about the line $x = 5$.
I get that in order for the equation to reflect about the $y$-axis, the function would have to be $y= - e^x$, and also for it to reflect about the $x$-axis the function should be something like $y= e^{-x}$.
But what to do when a line is not one of the coordinate axes?
Adding to @Adriano's answer, if you want to reflect $y=f(x)$ across $y=mx+b$ you have to reflect the function by the x-axis and then rotate it by $\theta=\pi+2\tan^{-1}(m)$ radians.
So if we have $y=f(-x)$, then by using the rotation matrix in linear algebra, we can set $x=x'\cos(\theta)+y'\sin(\theta)$ and $y=y'\cos(\theta)-x'\sin(\theta)$.
So the reflection over $y=mx$ equals
$$\left(\frac{2m}{2m^2+1}x'-\frac{1-m^2}{m^2+1}y'\right)=f\left(-\left(-\frac{2m}{m^2+1}y'-\frac{1-m^2}{m^2+1}x'\right)\right)$$
A better explanation for this formula is shown here.
Then with $b$, you have to shift $y=f(-x)$ down $b$ vertically to get $(y+7)=f(-x)$. Then after the rotation, the rotated function should be shifted up vertically by seven. Thus the final formula is
$$\left(\frac{2m}{2m^2+1}x'-\frac{1-m^2}{m^2+1}(y'-b)+b\right)=f\left(\frac{2m}{m^2+1}(y'-b)+\frac{1-m^2}{m^2+1}x'\right)$$
Now substitute $f(x)=e^{-x}$
Since we only know how to reflect about the $x$ or $y$ axis, we're going to have to reduce the given problem to a simpler problem. Notice that reflecting about the line $y = 4$ is equivalent to:
So we get: \begin{align*} \boxed{y = e^x} &\xrightarrow{~~~~\text{subtract }4~~~~~} \boxed{y = e^x - 4} \\ &\xrightarrow{\text{multiply by }-1} \boxed{y = 4 - e^x} \\ &\xrightarrow{~~~~~~~~~\text{add }4~~~~~~~~} \boxed{y = 8 - e^x} \\ \end{align*} See if you can do something similar for the other problem.
Use the reflection formula: $$\frac{x'-x}a=\frac{y'-y}b=\frac{-2(ax+by+c)}{a^2+b^2}\quad\text{for line }ax+by+c=0$$ Case A:
Line is $y-4=0$: $$\frac{x'-x}0=\frac{y'-y}1=\frac{-2(y-4)}{1}$$ where a convention is to take $0/0$ which is a result of the derivation. $$\implies x'=x,y'=8-y$$ So curve becomes: $$y'=8-e^{x'}\equiv y=8-e^x$$
Case B:
Line is $x-5=0$: $$\frac{x'-x}1=\frac{y'-y}0=\frac{-2(x-5)}{1}$$ where a convention is to take $0/0$ which is a result of the derivation. $$\implies x'=10-x,y'=y$$ So curve becomes: $$y'=e^{10-x'}\equiv y=e^{10-x}$$
Case C:
Let the line be $2x+3y+4=0$:
$$\frac{x'-x}2=\frac{y'-y}3=\frac{-2(2x+3y+4)}{13}$$
$$\implies x'=\frac1{13}(5x-12y-16), y'=\frac1{13}(-12x-5y-24)$$
Or:
$$x=\frac1{13}(5x'-12y'-16),y=\frac1{13}(-12x'-5y'-24)$$
So curve becomes:
$$\large (-12x'-5y'-24)=13e^{(5x'-12y'-16)/13}\\\large \equiv -12x-5y-24=13e^{(5x-12y-16)/13}$$
