The graph of the rational function \(\frac{p(x)}{q(x)}\) is shown below, with a horizontal asymptote of \(y = 0\). If \(q(x)\) is quadratic, \(p(2)=1\), and \(q(2) = 3\), find

\(p(x) + q(x).\)

Guest May 10, 2019

#1**+1 **

\(\text{Given the removable singularity at }x=1 \text{, and the vertical asymptote at }x=-1\\ \text{and the fact that }q(x) \text{ is quadratic}\\ q(x) =q_0 (x-1)(x+1)\\ \text{because of the removable singularity we also know that }p(x) \text{ has }(x-1) \text{ as a factor}\)

\(q(2)=3\\ q_0 (1)(3)=3\\ q_0=1\\ q(x) = (x+1)(x-1)=x^2-1\)

\(\text{As }\dfrac{p(x)}{q(x)} \text{ has a horizontal asymptote of }0\\ \text{the degree of }p(x) \text{ must be less than that of }q(x)\\ \text{so }p(x)=p_0(x-1)\\ p(2)=1\\ p_0=1\\ p(x)=x-1\)

\(p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2\)

.Rom May 11, 2019

#1**+1 **

Best Answer

\(\text{Given the removable singularity at }x=1 \text{, and the vertical asymptote at }x=-1\\ \text{and the fact that }q(x) \text{ is quadratic}\\ q(x) =q_0 (x-1)(x+1)\\ \text{because of the removable singularity we also know that }p(x) \text{ has }(x-1) \text{ as a factor}\)

\(q(2)=3\\ q_0 (1)(3)=3\\ q_0=1\\ q(x) = (x+1)(x-1)=x^2-1\)

\(\text{As }\dfrac{p(x)}{q(x)} \text{ has a horizontal asymptote of }0\\ \text{the degree of }p(x) \text{ must be less than that of }q(x)\\ \text{so }p(x)=p_0(x-1)\\ p(2)=1\\ p_0=1\\ p(x)=x-1\)

\(p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2\)

Rom May 11, 2019