# For the following function f (x) = 1/ -(x - 5)^2 + 7 determine: a) equation of the asymptote as exact valuesb) coordinates of the invariant points as exact values

rcmath | Certified Educator

Given the function `f(x)=1/-(x-5)^2+7`

To find find the horizpntal asymptote, we look at the limit when x tends towards infinity. We obtain `lim_(x->oo)f(x)=0+7=7`

Hence Horizontal Asymptote is y=7

As for the vertical asymptote, we notice that the function is not define for x=5 (Makes the denominator 0). That value yeild the vertical asymptote.

Vertical Asymptote is x=5

This function has no oblique asymptote.

oldnick | Student

`y=-1/(x-5)^2+7`

Now since  `lim_(x->5) -1/(x-5)^2 +7= oo`  then the  line `x=5`  is a vertical asymptote.

Indeed the fuction is defined for `x!=5`  while si domain is `(-oo;oo)`

Further :  `lim_(x->oo) -1/(x-5)^2 +7=7`   so that the line `y=7 ` is an horizzontal asymptote

Now we gonna see invaraint points

`y'= 2/(x-5)^3`

`y'!=0`  for every x.

Let you se as the  vertical line x=5  works as veretical asymptote, while the horizontal  line:y=7 (In blue) as horizontal one.