What is the slope and #y#-intercept in case of following cases representing lines?

#f(x)# passes through #(-2,2)# and also through #(0,-4)#
#g(x)# is given by #y=3x-4#
#h(x)# represents apples sold on Monday as #4# and on Tuesday, it is #1#
#j(x)# moves through points #(-2,-2)#, #(2,10)#, #(6,221)#

1 Answer
Shwetank Mauria
Sep 23, 2016

#f(x)# - slope is #-3# and #y#-intercept is #-4#
#g(x)# - slope is #3# and #y#-intercept is #-4#
#h(x)# - slope is #-3# and #y#-intercept is #4#
#j(x)# - slope is #3# and #y#-intercept is #4#

Explanation:

#f(x)# - This line passes through #(-2,2)# and #y#-intercept is #-4# i.e. it passes through #(0,-4)#. Hence its slope is #(-4-2)/(0-(-2))=-6/2=-3#. Hence #f(x)# is given by #y=-3x-4#

#g(x)# - As give this is given by #y=3x-4#, hence slope is #3# and intercept on #y#-axis is #-4#.

#h(x)# - here we assume that Monday is #0# and Tuesday is #1#, hence as number of apples on #0# day is #4# and on #1# day is #1#, the representative #y# intercept is #4# and slope is #(1-4)/(1-0)=-3/1=-3#.

#j(x)# - here as the #x# in function moves from #-2# to #2# and then #6# and #j(x)# moves from #-2# to #10# and then #22#, the slope is #(10-(-2))/(2-(-2))=12/4=3# also #(22-10)/(6-2)=12/4=3#. For #y#-intercept, we can take mid point of #2# and #-2#, where #j(x)=(10+(-2))/2=8/2=4# and #y#-intercept is #4#

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