In the graph of $\frac{x^2+3x}{x^2+4x+3}$, let $a$ be the number of holes in the graph, $b$ be the number of vertical asympotes, $c$ be the number of horizontal asymptotes, and $d$ be the number of oblique asymptotes. Find $a+2b+3c+4d$.
To analyze the graph of a rational function, I need to factor the numerator and denominator and look for any common factors that would cancel out and create holes in the graph. I can factor the numerator as $x(x+3)$ and the denominator as $(x+1)(x+3)$. I see that $x+3$ is a common factor, so that means there is a hole in the graph when $x=-3$. That's one hole, so $a=1$. To find the vertical asymptotes, I need to look for any values of $x$ that would make the denominator zero and are not holes. The only such value is $x=-1$, since $x=-3$ is a hole. That's one vertical asymptote, so $b=1$. To find the horizontal or oblique asymptotes, I need to compare the degrees of the numerator and denominator. The degree of both the numerator and denominator is 2, so they are the same degree. That means there is a horizontal asymptote at the ratio of the leading coefficients, which is $\frac{1}{1}=1$. That's one horizontal asymptote, so $c=1$. Since there is a horizontal asymptote, there cannot be any oblique asymptotes, so $d=0$. Now I have all the values I need to find $a+2b+3c+4d$. Plugging them in, I get $1+2(1)+3(1)+4(0)=1+2+3+0=6$. 