How do you find all the asymptotes for $f (x) = \frac{x^{2} - 5 x + 6}{x - 4}$ $f(x) =(x^2-5x+6)/(x-4)$ ?

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How do you find all the asymptotes for $f (x) = \frac{x^{2} - 5 x + 6}{x - 4}$ $f(x) =(x^2-5x+6)/(x-4)$ ?

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Answer 1 · 2015-07-15T19:31:21

If your studies have also included oblique (aka slant or skew) asymptotes, you'll need to find that also. The oblique asymptote is $y = x - 1$ $y = x-1$

Explanation:

Because the degree of the numerator is exactly $1$ $1$ more than the degree of the denominator, there is an oblique asymptote. To find it, divide $\frac{x^{2} - 5 x + 6}{x - 4}$ $(x^2 - 5x + 6)/(x-4)$

I don't have a great format for long division for Socratic, but this may help:

$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$

What do we need to multiply $x$ $x$ by, to get $x^{2}$ $x^2$ ? We need to multiply by $x$ $x$

$x$ $" " " " " " "x$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$

Now multiply $x$ $x$ times the divisor, $x - 4$ $x-4$ , to get $x^{2} - 4 x$ $x^2-4x$ and write that under the dividend.

$" " " " " " "$ $x$ $x$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$
$" " " " " "$ $x^{2} - 4 x$ $x^2-4x$
$" " " "$ $- - - - -$ $-----$

Now we need to subtract $x^{2} - 4 x$ $x^2-4x$ from the dividend. (Many find it simpler to change the signs and add.)

$" " " " " " "$ $x$ $x$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$
$" " " " " "$ $x^{2} - 4 x$ $x^2-4x$
$" " " "$ $- - - - -$ $-----$
$" " " " " "" "$ $- x$ $-x$ $+ 6$ $+6$

Now, what do we need to multiply $x$ $x$ (the first term of the divisor) by to get $- x$ $-x$ ? We need to multiply by $- 1$ $-1$

$" " " " " " "$ $x$ $x$ $- 1$ $-1$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$
$" " " " " "$ $x^{2} - 4 x$ $x^2-4x$
$" " " "$ $- - - - -$ $-----$
$" " " " " "" "$ $- x$ $-x$ $+ 6$ $+6$

Do the multiplication: $- 1 \times (x - 4)$ $-1xx(x-4)$ and write the result underneath:

$" " " " " " "$ $x$ $x$ $- 1$ $-1$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$
$" " " " " "$ $x^{2} - 4 x$ $x^2-4x$
$" " " "$ $- - - - -$ $-----$
$" " " " " "" "$ $- x$ $-x$ $+ 6$ $+6$
$" " " " " "" "$ $- x$ $-x$ $+ 4$ $+4$

Now subtract (change the signs and add), to get:

$" " " " " " "$ $x$ $x$ $- 1$ $-1$
$" " " "$ $- - - - -$ $-----$
$x - 4 ∣$ $x-4 |$ $x^{2}$ $x^2$ $- 5 x$ $-5x$ $+ 6$ $+6$
$" " " " " "$ $x^2-4x$
$" " " "$ $-----$
$" " " " " "" "$ $-x$ $+6$
$" " " " " "" "$ $-x$ $+4$
$" " " "$ $-----$
$" " " " " "" "" " " "$ $+2$

Now we see that:

$(x^2-5x+6)/(x-4) = x-1+2/(x-4)$ The difference (subtraction) between $y=f(x)$ and the line $y=x-1$ is the remainder term: $2/(x-4)$ .

As $x$ gets very very large, whether positive or negative, this difference gets closer and closer to $0$ . So the graph of $f(x)$ gets closer and closer to the line $y=x-1$

The line $y=x-1$ is an oblique aymptote for the graph of $f$ .

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How do you find all the asymptotes for $f (x) = \frac{x^{2} - 5 x + 6}{x - 4}$ $f(x) =(x^2-5x+6)/(x-4)$ ?

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Explanation:

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How do you find all the asymptotes for $f (x) = \frac{x^{2} - 5 x + 6}{x - 4}$ $f(x) =(x^2-5x+6)/(x-4)$ ?