# How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given f(x)=(x-4)^2/(x^2-4)?

Feb 7, 2017

Relative maximum $f \left(1\right) = - 3$ and relative minimum f(4)=0. Asymptotes : $\uparrow x = \pm 2 \downarrow . P O I : \left(5.034 , 0.0500\right) . f \downarrow$ : x in (1, 2) and (2, 4)$\mathmr{and}$f uarr :, x in (-oo, -2), (-2, 1) and (4, oo) 

#### Explanation:

I took 30' for this answer. I should have a break. Please make another answer, without attempting to edit my answer. After some time, I would edit my answer, myself.. I have to repeat this because some rush to supply typo omissions of characters like . I would do it myself. Logical and computational errors can be pointed out in the comment column and not by editing my answer. .

graph{(x-4)^2/(x^2-4) [-30, 30, -15, 15]}

x-intercept ( y = 0) : 4

y-intercept ( x = 0 ) : $- 4$

By actual division and resolving into partial fractions,

$y = f = 1 + \frac{1}{x - 2} - \frac{9}{x + 2}$

As $x \to 2 \mathmr{and} - 2 , f \to \pm \infty$

$f ' = - \frac{1}{x - 2} ^ 2 + \frac{9}{x + 2} ^ 2$

$= 8 \frac{\left(x - 1\right) \left(x - 4\right)}{{x}^{2} - 4} ^ 2$

=0, when x = 1 and 4,

$f \downarrow$ intervals:

$< 0 , x \in \left(1 , 2\right) \mathmr{and} \left(2 , 4\right)$ and

$f \uparrow$ intervals :

$> 0 , x \in \left(- \infty , - 2\right) , \left(- 2 , 1\right) \mathmr{and} \left(4 , \infty\right)$

Local minimum f(4)=0

Local maximum : $f \left(1\right) = - 3$

$f ' ' = \frac{2}{x - 2} ^ 3 - \frac{18}{x + 2} ^ 3 = 0$, when ${9}^{\frac{1}{3}} \left(x - 2\right) = \left(x + 2\right)$, giving

$x = 2 \frac{{9}^{\frac{1}{3}} + 1}{{9}^{\frac{1}{3}} - 1} = 5.7034 , \ne a r l y$

$f ' ' ' = - \frac{6}{x - 2} ^ 4 + \frac{54}{x + 2} ^ 4 \ne 0 , a t x = 5.034$, and here y = 0.05,

nearly

So, the POI is at x = 5.034, nearly