f(2x + 1) transformation

Question

I was working on a problem that asked:

If there is a vertical asymptote at $x = 5$ for $f(x)$, where is the vertical asymptote for $f(2x + 1)$?

The correct answer is at $x = 2$, but this confused me because the transformation would seem to suggest that the asymptote was at $5/2 - 1$, or $1.5$. Only when the $+ 1$ part of the transformation is done first does the correct answer of $x = 2$ result, but this seems to violate the order of operations. Can someone explain?

Answer 1

Note that it isn't the same $x$. Therefore, if for $x_0$ there is an asymptote for $f(x),f(x_0)$, then this same $f(x)$ at point $x_1=2 x_0 + 2$, there is also the asymptote. So $x_0 = 5$ and $x_1=2x_0+1$, then $x_1 = 2$.

Answer 2

Let $y=2x+1$. Then $f(y)$ has an asymptote at $y=5$ (given). So the asymptote of $f(y)$ is at $$ 2x+1=y=5. $$ Solving, we find $x=2$.

Answer 3

The problem just boils down to figuring out where $2x + 1 = 5$. You're just setting

\begin{align*}\text{input of } f &= \text{location of old asympote}\\ 2x + 1 &= 5 \end{align*}

since you know $f$ has an asymptote at $x = 5$.

Remember that you can solve an equation in any order you'd like; the order of operations only tells you how to do arithmetic.

You could solve using the sequence of steps

\begin{align*} 2x + 1 &= 5 \\ x + 1/2 &= 5/2 \tag{divide by 2}\\ x &= 5/2 - 1/2 = 2, \tag{subtract 1/2} \end{align*}

or by doing the usual thing and subtracting $1$ first, then dividing by $2$.

Answer 4

You can think of a vertical asymptote being a value of $x$ for which the function has a problem. Think of $\tan (90^{\circ})$ or $\frac 10$ - old calculators would say "Math error."

Your function has a problem when the input value $x$ is equal to $5$.

Your transformed function will have a problem when $2x+1=5$, which gives $x=2$. So the asymptote is now at $x=2$.

Answer 5

To make the function $f(2x+1)$ more analytical we need to express it as follows $$f(2(x+\frac{1}{2}))$$ This suggests that we perform the transformation by compressing horizontally by a factor of 2, and then shifting to the left by $\frac{1}{2}$. Compressing $5$ by a factor of $2$, we get $5/2$, and then shifting to the left by $1/2$ we get $2$.

This is easily seen by looking at a one-to-one portion of the function $f(2x+1)$, and turning it inside out (algebraically).

$$\begin{array}{lll} g(x)&=&f(2x+1)\\ f^{-1}(g(x))&=&2x+1\\ f^{-1}(g(x))-1&=&2x\\ \frac{1}{2}(f^{-1}(g(x))-1)&=&x\\ x&=&\frac{1}{2}f^{-1}(g(x)) - \frac{1}{2}\\ \end{array}$$ Now it is a bit easier to see why "$x$" becomes compressed and (in this form) shifted downwards (because we are now looking at $x$ on the vertical axis). That downward shift becomes a leftward shift when we place $x$ again on the horizontal axis.

Now lets go back the other way, this time, expressing $f$ in a more analytical form. $$\begin{array}{lll} x&=&\frac{1}{2}f^{-1}(g(x)) - \frac{1}{2}\\ x+\frac{1}{2}&=&\frac{1}{2}f^{-1}(g(x))\\ 2(x+\frac{1}{2})&=&f^{-1}(g(x))\\ f(2(x+\frac{1}{2}))&=&g(x)\\ g(x) &=& f(2(x+\frac{1}{2}))\\ \end{array}$$

Answer 6

You could also make up an example for a function that gives you an x asymptote of 5 to begin with. The most simple function would be

f(x) = 1/(x-5)

You now apply the transformation f(2x+1) into the original function.

You get f(2x+1) = 1/(2x+1-5)

f(2x+1) = 1/(2x-4)

Then solve for denominator = 0 since we can never divide by 0 to get your asymptote.

2x - 4 = 0

x = 2

Voila