What is the vertex form of #7y=3x^2+2x − 4.#?

1 Answer
Feb 4, 2016

#y=3/7(x+1/3)^2-13/21#
Please check the calculations!

Explanation:

write as:

#y=3/7x^2+2/7x-4/7#..................................(1)

#y=3/7(x^2+color(blue)(2/3x))-4/7#

consider the #2/3" from "color(blue)( 2/3x)" and multiply it by "color(brown)( 1/2)#

#color(brown)(1/2)xxcolor(blue)(2/3)=color(green)(1/3)#

#y!=3/7(x+color(green)(1/3))^2-4/7" "##color(purple)(" This introduces an error!")#

Let #k# be some constant then:

#y=3/7(x+1/3)^2+k-4/7# ...................(2) #color(purple)("Corrected the error!")#

expanding to find the value of k
#y=3/7x^2+2/7x+1/21+k-4/7# ......................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Equate equation (1) to equation (3) through y

#cancel(3/7x^2)+cancel(2/7x)-cancel(4/7)" "=" "cancel(3/7x^2)+cancel(2/7x)+1/21+k-cancel(4/7)#

#k+1/21=0" "->" "k=-1/21#.............................(4)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (4) into (2) giving

#y=3/7(x+1/3)^2-1/21-4/7 ...................(2_a)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solution:

#y=3/7(x+1/3)^2-13/21#

#color(purple)("Please check the arithmetic. I can not spot any error but I am not")# #color(purple)("totally satisfied with the answer!")#