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
