Answer
See explanation
Work Step by Step
Given
$$
y=a b^x
$$ $y'=ab^x\ln b$
$y''=ab^x(\ln b)^2$
As $b^x>0$ for all $x$, the sign of $y'$is the same as the sign of $a\ln b$ and the sign of $y''$ is the same as the sign of $a$ since $(\ln b)^2\geq 0$ and is $0$ only when $b=1$.
So $y$ is increasing when $a>0$ and $b>1$ or when $a<0$ and $00$ and $01$.
$y$ is concave up on $\mathbb{R}$ when $y''>0$, i.e. $a>0$ and $b\not=1$.
$y$ is concave down on $\mathbb{R}$ when $y''<0$, i.e. $a<0$ and $b\not=1$.
If $a=0$, then $y=0$ (constant) (neither increasing, nor decreasing).
If $b=1$, then $y=a$ (constant) (neither increasing, nor decreasing).
