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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 $x = 5$.

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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?

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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:

- Translating $4$ units down.
- Reflecting about the $x$-axis.
- Translating $4$ units up.

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:

$$\large (-12x’-5y’-24)=13e^{(5x’-12y’-16)/13}\\\large \equiv -12x-5y-24=13e^{(5x-12y-16)/13}$$

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 $f(x)=e^{-x}$

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