Pretty sure I got this right, but just want confirmation.
Find the constants a and b such that the function is continuous everywhere.
$$f(x)=\begin{cases} {5x-2},&\text{if }x<1\\\\ a,&x=1\\\\ ax^2+bx&\text{if }x> 1\; \end{cases}$$
For the function to be continuous, the left and right sided limits have to be equal. In this case, we are approaching x=1 from the left and right. I find the left side limit, f(x)=5x-2 as 3. The limit f(1)=a would therefore have to equal 3, which means a=3. Which means the right side limit f(x)=ax^2+bx has to equal 3, and if a=3, then b=0.
Yes? No?