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#
