If you don't remember the exact distribution for $Y=a+bX$, you can always derive it. First, suppose $b>0$. Then,
\begin{aligned}
\Pr[Y\leq y]&=\Pr[a+bX\leq y]\\
&=\Pr[X\leq (y-a)/b]\\
&=\int_{-\infty}^{(y-a)/b}\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{1}{2}x^2\right]dx
\end{aligned}
You desire the upper limit to be $y$ so make the substitution $u=a+bx$. As $x$ varies between $-\infty$ and $(y-a)/b$, $u$ varies between $-\infty$ and $y$ (this is where we use $b>0$). We have
$$
\Pr[Y\leq y]=\int_{-\infty}^y\frac{1}{b\sqrt{2\pi}}\exp\left[-\frac{1}{2}\frac{(u-a)^2}{b^2}\right]du.
$$
You then can recognize the integrand as the PDF of $N(a,b^2)$.
If $b<0$, then $Y=a+bX=a+(-b)Z$, where $Z=-X$ is also $N(0,1)$ (why?). By the analysis above, $Y\sim N(a,(-b)^2)=N(a,b^2)$.
Finally, if $b=0$, then $Y$ is nonrandom and always equals $a$, which can be thought of as $N(a,0)=N(a,b^2)$.
So in all cases, you have $Y\sim N(a,b^2)$.