### 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 $3x^2+5x+2=0$.
This is a quadratic equation, so I can use the quadratic formula to solve for $x$.
Alternatively, I could try to factor the quadratic, but it doesn't look like it has any obvious integer factors, so I'll use the formula.
The quadratic formula says that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
So in this case, $a=3, b=5, c=2$, and I get $x=\frac{-5\pm\sqrt{25-24}}{6}$.
Simplifying, I get $x=\frac{-5\pm\sqrt{1}}{6}=\frac{-5\pm 1}{6}$.
So the two values of $x$ that make the denominator zero are $x=\frac{-4}{6}=\frac{-2}{3}$ and $x=\frac{-6}{6}=-1$.
These are the values of $x$ where the vertical asymptotes are, so $a=\frac{-2}{3}$ and $b=-1$.