Question 1034488
{{{f(x) = x^3 +x^2 +x+1}}}  ==> f'(x) = {{{3x^2+2x+1}}}.
Now the line  2y + x +5=0 has slope -1/2.  Any line perpendicular to it must have slope 2.  
We now find points on the graph whose tangents have slope 2.

==> f'(x) = {{{3x^2+2x+1 = 2}}} ==>  {{{3x^2+2x-1 = 0}}}

==> (3x-1)(x+1) = 0 ==> x = 1/3, x = -1.

Now f(1/3) = 40/27, and f(-1) = 0.

==> The two lines are {{{y-40/27 = 2(x-1/3)}}} and {{{y-0 = 2(x+1)}}}.

Simplify these equations yourself.