### Instruction: The graph of $y=\frac{5x^2-9}{3x^2+5x+2}$ has vertical asymptotes at $x = a$ and $x = b$.  Find $a + b$.### Response: I know that vertical asymptotes occur when the denominator of a rational function is equal to zero, so I want to find the values of $x$ that make $3x^2+5x+2=0$.
To do that, I can use the quadratic formula, which says that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $a=3$, $b=5$, and $c=2$, so I plug those values into the formula and simplify.
I get $x=\frac{-5\pm\sqrt{25-24}}{6}=\frac{-5\pm 1}{6}$.
So the two values of $x$ that make the denominator zero are $x=\frac{-5+1}{6}=-\frac{2}{3}$ and $x=\frac{-5-1}{6}=-1$.
These are the values of $a$ and $b$, so $a+b=-\frac{2}{3}-1=-\frac{5}{3}$.