First method. The graph intersects the y-axis at the point $(0, b)$. Therefore, $b>0$, which means $|b|=b$. When $x=1$, $y=k+b$, and from the graph, it is clear that $0-k=|k|$.

Second method. Let $(a, 0)$ be the point where the graph intersects the x-axis. From the graph, it is clear that $a>1$. Therefore, $|k|=|b|:|a|<|b|$.

## Answer

$|k|<|b|$.
Folklore

The graph of the function $y=kx+b$ is shown in the figure. Compare $|k|$ and $|b|$.

![](https://cdn.mathpix.com/cropped/2024_05_06_85fcd68678dc231c8adbg-08.jpg?height=389&width=323&top_left_y=1632&top_left_x=868)