How do you sketch the curve `y=(x+1)/sqrt(5x^2+35)` by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

`f(x)=(x+1)/(sqrt(5x^2+35))`

Vertical asymptote is in the point where it isn't defined that is:
`sqrt(5x^2+35)!=0`
`5x^2+35!=0`
`x^2!=-35/5`
`x!=sqrt(-35/5)`
As you can see this is absolute nonsense. That can't be done. Square root must be `>0`. That means it doesn't have a vertical asymptote and The Domain is `RR`

In order to determine maximum and minimum we need first derivative:
`f(x)=(x+1)/(sqrt(5x^2+35))`

`f^'(x)=((x+1)^'(sqrt(5x^2+35))-(x+1)((5x^2+35)^(1/2))^')/(sqrt(5x^2+35))^2`

`f^'(x)=(1*sqrt(5x^2+35)-(x+1)((1*10x))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2`

`f^'(x)=((2(sqrt(5x^2+35))^2-10x(x+1))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2`

`f^'(x)=(2(5x^2+35)-10x(x+1))/(2(sqrt(5x^2+35))^3`

`f^'(x)=(cancel(10x^2)+2*35cancel(-10x^2)-2*5x)/(2(sqrt(5x^2+35))^3`

`f^'(x)=(cancel2(35-5x))/(cancel2(sqrt(5x^2+35))^3`

`f^'(x)=(5(7-x))/(sqrt(5x^2+35))^3`

`sqrt(5x^2+35)` is always positive so we don't care about that. However `(7-x)` can be + and -
`x in (-oo,7) hArr f^'(x) >0 =>` f goes up
`x in (7,oo) hArr f^'(x) <0 =>` f goes down

maximum is in x=7`quadf(7)=(7+1)/(sqrt(5*(7)^2+35))=8/(sqrt(280))=4/sqrt70~~0.478`

function doesn't have minimum

horizontal asymptote: `Lim_(xrarr+-oo)f(x)`
1. for `+oo`
`Lim_(xrarroo)(cancelx(1+1/x))/(cancelx(sqrt(5+35/(x^2))))=Lim_(xrarroo)(1+1/x)/(sqrt(5+35/(x^2)))=(1+0)/(sqrt(5+0))=1/sqrt5~~0.447`

1. for `-oo`
   `Lim_(xrarr-oo)f(x)="by implementing L'Hopitals rule we get"=Lim_(xrarr-oo)(sqrt(5x^2+35)*5x)=-oo`

Intercepts:
`if x=0quad=>quad y=1/sqrt35`

`if y=0quad=>quad 0=x+1quad=>quadx=-1`