Question 269403
use the equation y=mx+b where m=slope and b=y-intercept
{{{y=(-5/4)x+1}}}
{{{m=(-5/4)}}} slope = y-axis/x-axis
{{{b=1}}}
im not going to show the graph since your asking how to graph
<p>
starting at point (0,1) and with the given slope(-5/4) the easiest way to find the otherpoints is to use this plot system where a/b=slope 
(x+b,y+a)
(x+(4),y+(-5))
(0+(4),1+(-5))
(4,-4)
(x+b,y+a) repeat with the previous point
(x+(4),y+(-5))
(4+(4),-4+(-5))
(8,-9)
now you have 3 points to draw a line
<p>
OR
<p>
if you cant remember how to use the slope to make your points then you will have to find points to make your line.
<p>
easiest way to draw the line is to make a chart of x-coordinates that will give u easy y-coordinates to graph. normally i would do {x=0,1} but since we are dealing with a fraction for the slope, i would want to use numbers that would cross cancel and leave whole numbers. I would use {x=4,-4}. i will use these numbers as they are easily divisiable by 4 which is the denominator of the slope. so lets find our y-cordinates:
{{{y=(-5/4)x+1}}}
{{{y=(-5/4)4+1}}} and {{{y=(-5/4)(-4)+1}}}
we get these two pairs: and we know the line crosses y-axis at +1.
(4,-4)
(-4,6)
plot the points and draw your line
(-4,6),(0,1),(4,-4)
should look like a "\" line