To find the #x# intercept, we must set #y = 0# and solve for #x#

#0=-4-4x-> 4=-4x->-1=x#

We found that #x=-1# so our #x#-intercept is at #(-1,0)#

Note: The #x#-intercept is where our function crosses the #x#-axis

To find the #y#-intercept, we must set #x=0# and solve for #y#

#y=-4-4(0)->y=-4#

We found that #y=-4# so our #x#-intercept is at #(0,-4)#

Note: The #y#-intercept is where our function crosses the #y#-axis. Also, we didn't have to find the #y#-intercept this way because In any linear function the #y#-intercept is always the constant (or #color(red)b#) in the equation #y=mx+color(red)b#

Now that we have both s our #x# & #y# intercepts we can graph this (see graph).

Note that our slope is #-4# since the slope is the coefficient (or #color(blue)m#) in the equation #y=color(blue)mx+b#

I hope this was helpful and that you understand the topic a little better now!