With [tex]f[/tex] defined by
[tex]f(x)=\begin{cases}|x-1|+2&\text{for }x<1\\ax^2+bx&\text{for }1\le x<2\\5x-10&\text{for }x\ge2\end{cases}[/tex]
in order for it to be continuous at [tex]x=c[/tex], we require
[tex]\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)[/tex]
(i) If [tex]a=2[/tex] and [tex]b=3[/tex], then [tex]f(1)=2(1)^2+3(1)=5[/tex] and
[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}|x-1|+2=2[/tex]
[tex]\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}2x^2+3x=5[/tex]
The limits don't match, so [tex]f[/tex] is not continuous at [tex]x=1[/tex] under these conditions.
(ii) To establish continuity at [tex]x=1[/tex], we'd need the limit as [tex]x\to1[/tex] from the right to be equal to the limit from the left, or
[tex]\displaystyle\lim_{x\to1}ax^2+bx=\lim_{x\to1}|x-1|+2\iff a+b=2[/tex]
(iii) We have [tex]f(2)=0[/tex] and
[tex]\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}ax^2+bx=4a+2b[/tex]
[tex]\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}5x-10=0[/tex]
For [tex]f[/tex] to be continuous at [tex]x=2[/tex], then, we'd need to have
[tex]4a+2b=0[/tex]
(iv) Taking both requirements from parts (ii) and (iii), we solve for [tex]a,b[/tex]:
[tex]\begin{cases}a+b=2\\4a+2b=0\end{cases}\implies a=-2,b=4[/tex]
I've attached a plot that confirms this is correct.
