Question

How do you find the important points to graph `f(x)= -x^2-4x`?

Answer 1

Check below for detail examination of the function.

Explanation:

`f(x)=-x^2-4x` ,
`D_f=RR`

`f(x)=0 <=> -x^2-4x=0 <=> x^2+4x=0 <=> x(x+4)=0 <=> (x=0, x=-4)`

`f'(x)=-2x-4=-2(x+2)`

`f'(x)=0` `<=> x=-2`

`f''(x)=-2<0`
`x` `in` `RR`

• We get these tables for monotony and concavity of `f`:
  

`f` has global maximum at `x_0=-2` , `f(-2)=4`

`lim_(xrarr-oo)f(x)=lim_(xrarr-oo)(-x^2-4x)=lim_(xrarr-oo)(-x^2)=-oo`

`f(-2)=4`

`lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(-x^2-4x)=lim_(xrarr+oo)(-x^2)=-oo`

• As a result the range of `f` will be `(-oo,4]`

`f` is defined in `RR` so it doesn't have any vertical asymptotes.

`lim_(xrarr-oo)f(x)/x=lim_(xrarr-oo)(-x^2-4x)/x=lim_(xrarr-oo)(-x^2/x)=+oo`

• As a result `f` doesn't have any oblique/horizontal asymptotes.