<s>[INST] Find values of $a$ and $b$ that make the function continuous everywhere. I need some help with this question:

Find the values of $a$ and $b$ that make $f$ continuous everywhere.
$$f(x)=\begin{cases}
x^2 − 4/x-2, &\text{if }x < 2\\
ax^2-bx+1, &\text{if } 2 ≤ x ≤ 3\\
4x - a + b, &\text{if } x ≥ 3\end{cases}$$

I started by writing two expressions for $a$ and $b$ based upon the left and right limits of each piece's endpoints. But I ended up with $a=3/2$ and $b=2/3$. Where did I go wrong?
 [/INST] If $f$ is continuous then :
$f(2)=\underset{x\rightarrow2}{\lim}f(x)=2^2-\frac{4}2-2=a(2^2)-b(2)+1$
Therefore $4a-2b+1=4-2-2=0$ hence $4a-2b+1=0$.
Same reasoning around $x=3$ : 
$a(3^2)-b(3)+1=4(3)-a+b$ (ie) $9a+3b+1=12-a+b$ therefore $10a-4b=11$
You just need to solve $\begin{cases}4a-2b=-1\\10a-4b=11\end{cases}$ to find $a$ and $b$.
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