# 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

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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.**

`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.