Answer
See the explanation
Work Step by Step
As $r$ increases, the graph of $y=a(1+r)^t$ rises steeply, so the point of intersection moves to the left and down. However, no matter how steep the graph becomes, the point of intersection remains above and to the right of the $y$-intercept of the second curve, or the point $(0, b)$. Hence, the value of $y_0$ decreases but does not reach $b$.