How do you write a function rule for #x = 2, 4, 6# and #y = 1, 0, -1#?

1 Answer
Mar 20, 2018

Answer:

#y=f(x) = -1/2x+2#

Explanation:

Set the #i^("th")# point as: #P_i->(x_i,y_i)#

Change in #x# sequence is #2#
Change in #y# sequence is #-1#

Gradient (slope) #->m=("change in "y)/("change in " x) =-1/2#

Assuming there is a direct link between #x and y# numbers of sequence we have:

#P_1->(x_1,y_1)=(2,1)#
#P_2->(x_2,y_2)=(4,0)#
#P_3->(x_3,y_3)->(6 ,-1 )#

Relating #m# to, say, point 2 we have:

#m=-1/2=(y-y_2)/(x-x_2)=(y-0)/(x-4)#

#-1/2= (y-0)/(x-4)#

Multiply both sides by #(+2)#

#-1=(2(y-0))/(x-4)#

Multiply both sides by #(x-4)#

#-(x-4)=2y#

#-x+4=2y#

#y=-1/2x+2#
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The question is specific in that it states 'function rule for #x#'

So we write it as: #y=f(x) = -1/2x+2#

Tony B