Exercises: Linear functions and line equations

The graph is rising. So only options $2.5$ and $1$ can be correct. If you go from the $y$-axis intercept (here $y=-2.5$) to the right by $1$, you still have to go up by about $1$to reach the straight line again.

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

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

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

The line is falling. Therefore, the slope can only be negative. The only possible correct solutions are $0.8$ or $-1.2$.

If you move $1$ to the right in the coordinate system from the $y$-axis intercept, you have to move less than $1$ down to meet the straight line again. So the answer $-1.2$ cannot be correct.

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

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

The graph is rising, so the slope can only be positive. The answer options $2$ or $1.8$ or $2.2$ are therefore sensible at first glance.

Find a point on the straight line whose coordinates you can read off easily. For example, the point $(3|0)$ is suitable here. From here you have to go $1$ to the right and less than $2$ upwards to reach the straight line again. Therefore, only the answer option $m=1.8$ makes sense.

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

You start from the $y$-axis intercept (here $y=3$) and go $7$ to the right and $3$ downwards.

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