+0

# help

0
157
1

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

$$p(x) + q(x).$$

May 10, 2019

#1
+6046
+1

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

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

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

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

.
May 11, 2019

#1
+6046
+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