Answer
$y=-\dfrac{1}{4}x$ (red graph)
$y=-\dfrac{1}{4}x+2$ (blue graph)
Work Step by Step
Linear equations of the form $y=mx+b$ has a $y$-intercept of $b$ and a slope of $m.$ Hence, the given equation, $
y=-\dfrac{1}{4}x
$ has
\begin{align*}
y\text{-intercept: }&
0
\\\text{Slope: }&
-\dfrac{1}{4} \text{ or } \dfrac{-1}{4}
.\end{align*}
To graph the equation, start at the $y-$intercept of $(
0,0
)$. Then use the notion of the slope as $\dfrac{rise}{run}.$ Since the slope is $
\dfrac{-1}{4}
,$ then $rise=
-1
$ and $run=
4
.$ Hence, from the $y$-intercept, go down by $rise=
1
$ unit then go to the right by $run=
4
$ units. This results to the point $(
4,-1
)$. By connecting this point and the $y$-intercept, the graph of $
y=-\dfrac{1}{4}x
$ (red graph) is determined.
Similarly, the other given equation, $
y=-\dfrac{1}{4}x+2
$ has
\begin{align*}
y\text{-intercept: }&
2
\\\text{Slope: }&
-\dfrac{1}{4} \text{ or } \dfrac{-1}{4}
.\end{align*}
The $y-$intercept is $(
0,2
)$. Using the notion of the slope as $\dfrac{rise}{run},$ then $rise=
-1
$ and $run=
4
.$ Hence, from the $y$-intercept, go down by $rise=
1
$ unit then go to the right by $run=
4
$ units. This results to the point $(
4,1
)$. By connecting this point and the $y$-intercept, the graph of $
y=-\dfrac{1}{4}x+2
$ (blue graph) is determined.
