The graph is rising. So only options $2.52.5$ and $11$ can be correct. If you go from the $yy$-axis intercept (here $y=-2.5y=-2.5$) to the right by $11$, you still have to go up by about $11$to reach the straight line again.

The slope is therefore: $m={\displaystyle \frac{1}{1}}=1m=\backslash dfrac\{1\}\{1\}=1$.

You look for two points in the coordinate system whose coordinates you can read off easily. Here, for example, $(1\mathrm{\mid }3)(1|3)$ and $(3\mathrm{\mid }0)(3|0)$. To get from $(1\mathrm{\mid }3)(1|3)$ to $(3\mathrm{\mid }0)(3|0)$, you have to go $11$ to the right and $33$ down.

The slope is therefore: $m=\frac{-3}{2}=-1.5m=\backslash frac\{-3\}\{2\}=-1.5$

The line is falling. Therefore, the slope can only be negative. The only possible correct solutions are $0.80.8$ or $-1.2-1.2$.

If you move $11$ to the right in the coordinate system from the $yy$-axis intercept, you have to move less than $11$ down to meet the straight line again. So the answer $-1.2-1.2$ cannot be correct.

If you look at the graph very very carefully, you get the result:

$m=\frac{-0.8}{1}=-0.8m=\backslash frac\{-0.8\}\{1\}=-0.8$

The graph is rising, so the slope can only be positive. The answer options $22$ or $1.81.8$ or $2.22.2$ are therefore sensible at first glance.

Find a point on the straight line whose coordinates you can read off easily. For example, the point $(3\mathrm{\mid }0)(3|0)$ is suitable here. From here you have to go $11$ to the right and less than $22$ upwards to reach the straight line again. Therefore, only the answer option $m=1.8m=1.8$ makes sense.

The slope is therefore: $m=\frac{1.8}{1}=1.8m=\backslash frac\{1.8\}\{1\}=1.8$

You start from the $yy$-axis intercept (here $y=3y=3$) and go $77$ to the right and $33$ downwards.

The slope is therefore: $m={\displaystyle \frac{-3}{7}}=-{\displaystyle \frac{3}{7}}m=\backslash dfrac\{-3\}\{7\}=-\backslash dfrac\{3\}\{7\}$