[Math] Reflecting an exponential function over a y = 3 line.

exponential functionfunctionsreflectiontransformation

How would you write the equation of $f(x) = 4^x$ that reflects over the line $y = 3$?

I've put in $f(x) = 3 + 4^{-x}$ which I thought was the right answer, but it isn't.

Thanks in advance!

Best Answer

Hint: Reflecting around $y=3$ sends $y$ to $6-y$

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Functions – How to Reflect a Function About a Specific Line

Your idea will work if you just carry it fully through. First shift three units to the left, so the line of reflection becomes the y axis, then flip, and finally remember to shift three units back to the right to put the center line back where it belongs.

(This gives the $f(6-x)$ solution you already know).

[Math] reflecting a function about a line

We can implement this as a translation of the line to the origin, a reflection about the line through the origin, and then a translation back in the same direction.

In particular: suppose we want to reflect a point $(x_0,y_0)$ across this line.

First translate it to the point $(x_1,y_1) = (x_0,y_0 - b)$.
Then, reflect $(x_1,y_1)$ across the line $y = ax$ to get $$ (x_2,y_2) = \frac1{a^2 + 1} ([1 - a^2]x_1 + 2a\,y_1,2a\,x_1 + [a^2 - 1]y_1) $$
Finally, translate back to get $(x_3,y_3) = (x_2,y_3 + b)$

So, the curve parametrized by $$ x = t\\ y = f(t) $$ Becomes the curve parametrized by $$ x = \frac{1 - a^2}{a^2 + 1}\,t + \frac{2a}{a^2 + 1}(f(t) - b) \\ y = \frac{2a}{a^2 + 1}\,t + \frac{a^2 - 1}{a^2 + 1}(f(t) - b)+b $$ Note: there is no guarantee that we can write $y$ as a function of $x$.

Interesting cases: if we take $g(x) = 0x + b$, then we get $$ y = - (f(t) - b) + b = 2b - f(t)\\ x = t $$ which is simply the curve $y = 2b - f(x)$.

If we take $g(x) = x$, then we get $$ y = t\\ x = f(t) $$ This is the curve $x = f(y)$. When $f$ is invertible, we can rewrite this as $y = f^{-1}(x)$.

Best Answer

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Functions – How to Reflect a Function About a Specific Line

[Math] reflecting a function about a line

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