First we need to find the slope at the given point so we take the derivative of #f(x)#: #d/dx[f(x) = -x/3] -> f'(x) = -1/3#
This tells us the slope is #-1/3# at any given point, now that we have a slope we need to find a point. The question tells us our #x# value is #-3# so plugging that value into #f(x)# gives us #1#.
So our point is #(-3,1)# and our slope is #-1/3# now we can put this into slope point from which is:
#y-y_1 = m(x-x_1)# where #m# is slope and #(x_1, y_1)#
Plugging all values gives us:
#y-1 = -1/3(x- (-3))# which is also #y=-x/3#, we get the original equation because the graph is linear and so the function can represent the value at any given point since the slope does not change
