How do you graph #y + 1/2 = -1/3(x+1/2)#?

1 Answer
Feb 16, 2016

#y=x^2/3+x/3-5/12#

Explanation:

You need to transform the given equation into a more workable format.

Subtract #color(blue)(1/2) # from both sides.

#y+1/2color(blue)(-1/2)=-1/3(x+1/2)^2color(blue)(-1/2)#

#y+0=1/3(x+1/2)^2-1/2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Square the brackets

#y=1/3(x^2+x+1/4)-1/2#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply the contents of the brackets by #1/3#

#y=(x^2)/3+x/3+1/12 -1/2#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Simplify the constant

#y=x^2/3+x/3-5/12#
'~~~~~~~~~~~~~~~~~~~~~~~~~~
Produce a table of values and plot your graph

Tony B