Given: #y= (2x+7)(3x-1)" [1]"#

The vertex form of a parabola of this type is:

#y = a(x-h)^2+k" [2]"#

We know that the "a" in the vertex form is the same as the coefficient #ax^2# in standard form. Please observe the product of the first terms of the binomials:

#2x * 3x = 6x^2#

Therefore, #a = 6#. Substitute 6 for "a" into equation [2]:

#y = 6(x-h)^2+k" [3]"#

Evaluate equation [1] at #x = 0#:

#y= (2(0)+7)(3(0)-1)#

#y= 7(-1)#

#y= -7#

Evaluate equation [3] at #x=0 and y = -7#:

#-7 = 6(0-h)^2+k#

#-7 = 6h^2+k" [4]"#

Evaluate equation [1] at #x = 1#:

#y= (2(1)+7)(3(1)-1)#

#y= (9)(2)#

#y= 18#

Evaluate equation [3] at #x=1# and #y = 18#:

#18 = 6(1-h)^2+k#

#18 = 6(1-2h +h^2)+k#

#18 = 6-12h +6h^2+k" [5]"#

Subtract equation [4] from equation [5]:

#25 = 6-12h#

#19=-12h#

#h = -19/12#

Use equation [4] to find the value of k:

#-7 = 6h^2+k#

#k = -6h^2-7#

#k = -6(-19/12)^2-7#

#k = -529/24#

Substitute these values into equation [3]:

#y = 6(x--19/12)^2-529/24#