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# how to solve

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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.

Jun 24, 2021

#1
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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

Jun 24, 2021
#2
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As follows: Jun 24, 2021
edited by Alan  Jun 24, 2021
edited by Alan  Jun 24, 2021