`f(x)=(7x-5)/(x+6)`

Let's start with I and J

I) Horizontal Asymptote

`lim_(x->oo)f(x)=7/1=7`

Hence y=7 is the horizontal asymptote

J) Vertical Asymptote

`lim_(x->-6)f(x)=oo`

Hence x=-6 vertical asymptote.

`f'(x)=[7(x+6)-(7x-5)*1]/(x+6)^2=>`

`f'(x)=[7x+42-7x+5]/(x+6)^2=47/(x+6)^2`

A) Since f(x) is not defined at x=-6, then we have no critical points.

B & C) f'(x) is always positive, hence it is always increasing.

`f''(x)=[-47*2*(x+6)]/(x+6)^4=-94/(x+6)^3`

D & E) no local max or min

F) f(x) is concave up when f''(x)>0=>`(x+6)<0=>x<-6`

`(-oo,-6)`

G) f(x) is concave down on `(-6,oo)`

H) No inflection point.

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