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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.
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'!=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.
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