# How do you graph f(x)=(x-2)/(x-4) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jan 25, 2018

Undefined at $x = 4$
See explanation for the rest

#### Explanation:

$\textcolor{b l u e}{\text{Hole - undefined}}$

Hole is where the denominator becomes 0. The function becomes undefined at that point. So for this condition we have $x = 4$

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Lets consider the behaviour close to $x = 4$

$\textcolor{b l u e}{\text{Vertical Asymptot } \to + \infty}$

If $x = 4 + \delta 4$ where $\delta 4 > 0$ and very small then we have

$\frac{4 + \delta 4 - 2}{\cancel{4} + \delta 4 - \cancel{4}} \to \frac{2}{\delta 4} + 1$

${\lim}_{\delta 4 \to {\textcolor{w h i t e}{}}^{+} 0} \frac{2}{\delta 4} + 1 \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd")->color(white)("dd}} + \infty + 1 = + \infty$

$\textcolor{b l u e}{\text{Vertical Asymptot } \to - \infty}$

If $x = 4 - \delta 4$ where $\delta 4 > 0$ and very small then we have

$\frac{4 - \delta 4 - 2}{\cancel{4} - \delta 4 - \cancel{4}} \to \frac{2 - \delta 4}{- \delta 4} = - \frac{2}{\delta 4} + 1$

${\lim}_{\delta 4 \to {\textcolor{w h i t e}{}}^{+} 0} - \frac{2}{\delta 4} + 1 \textcolor{w h i t e}{\text{dd") ->color(white)("dd")kcolor(white)("dd")->color(white)("dd}} - \infty + 1 = - \infty$
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$\textcolor{b l u e}{\text{Horizontal Asymptot } \to + \infty}$

$\frac{x - 2}{x - 4}$

As $x > 0$ becomes increasing greater and greater the influences of the -2 and -4 become less and less significant. This continues until we have

${\lim}_{x \to + \infty} \frac{x - 2}{x - 4} \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd}} \frac{\infty}{\infty} = + 1$

As $x < 0$ becomes increasing less and less the influences of the -2 and -4 become less and less significant. This continues until we have

${\lim}_{x \to - \infty} \frac{x - 2}{x - 4} \textcolor{w h i t e}{\text{dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd}} \frac{- \infty}{- \infty} = + 1$