# For what values of x, if any, does f(x) = 1/(x^2-4x+4)  have vertical asymptotes?

Feb 6, 2016

$x = 2$.

#### Explanation:

To find the vertical asymptotes we have to find for which values the denominator is equal to 0. Simply set the polynomial on the denominator to 0 then solve for $x$:

${x}^{2} - 4 x + 4 = 0$
$\left(x - 2\right) \left(x - 2\right) = 0$
$\therefore x = 2$

Thus there is one vertical asymptote at $x = 2$.

Here is a graph of the function, the vertical asymptote is clearly indicated by a steep incline towards $x = 2$.