Let f(x) = (3x^4+3x^3+3x^2+3)/(x^2+2x-4). Find a polynomial g(x) so that the graph of f(x) + g(x) has a horizontal asymptote of y = 0.
+171 y=0 basically means the x axis.
if it wasnt for the\( f(x)+g(x)\) condition, you could just chuck a negative sign in front of it and you were good to go.
anyway -- we are given that \( f(x) = \frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) } \) -- now what is the value of the polynomial g(x) so that it satisfies the condition? lets set up this 'equation' :
\( \frac{\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}+g(x)=\frac{-\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}\)
\(g(x)=-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4}-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4} \)
\(g(x)=-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4} \)
lets confirm the condition:
\( f(x)+g(x)\) \(\Rightarrow\) \(\frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) } +-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4} = \boxed{\frac{-3x^4-3x^3-3x^2-3}{x^2+2x-4}}\)
if you graph it you would get:
: 
here is the link for it as well: https://www.desmos.com/calculator/v2w0ukjkoq
:D
