At what value of $a$ do the graphs of $y=ax^2+3x+1$ and $y=-x-1$ intersect at exactly one point?
To find the value of $a$ that makes the graphs intersect at one point, I need to find the solution to the equation $ax^2+3x+1=-x-1$. This is a quadratic equation, so I can try to solve it by bringing everything to one side and using the quadratic formula. I get $ax^2+4x+2=0$. To use the quadratic formula, I need to identify the coefficients $A=a$, $B=4$, and $C=2$. The formula is $x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$. I want this equation to have exactly one solution, which means the discriminant $B^2-4AC$ must be zero. So I set $B^2-4AC=0$ and plug in the values of $A$, $B$, and $C$. I get $4^2-4a(2)=0$. Simplifying, I get $16-8a=0$. Solving for $a$, I get $a=2$. This is the value of $a$ that makes the graphs intersect at one point. 