For-a-cubic-function-in-the-form-f-x-ax-3-bx-2-cx-d-What-must-be-true-of-a-b-c-and-d-in-order-for-the-function-to-be-able-to-be-converted-to-the-form-f-x-a-x-h-3-k-

Question Number 114253 by Dat_Das last updated on 18/Sep/20

For a cubic function in the form :

f (x) = {a x}^{3} + {b x}^{2} + c x + d

What must be true of a, b, c, and d in

order for the function to be able to be

converted to the form :

f (x) = a {(x - h)}^{3} + k

Answered by 1549442205PVT last updated on 18/Sep/20

We need must find a, b, c, d such that

{a x}^{3} + {b x}^{2} + c x + d = a {(x - h)}^{3} + k

\Leftrightarrow {a x}^{3} + {b x}^{2} + c x + d \equiv {ax}^{3} - {3 ahx}^{2} + {3 ah}^{2} x - {ah}^{3} + k

\Leftrightarrow {\begin{cases} b = - 3 ah \\ c = {3 ah}^{2} \\ d = - {ah}^{3} + k \end{cases}

where a is arbitrary; h, k given

Commented by Dat_Das last updated on 18/Sep/20

This is true, however you may relate

a, b, c directly . Take

b = - 3 a h

\frac{b}{- 3 a} = h

\frac{b^{2}}{9 a^{2}} = h^{2}

\frac{b^{2}}{3 a} = 3 {a h}^{2}

Substituting c for 3 {a h}^{2} we get

\frac{b^{2}}{3 a} = c

Therefore for any function

{a x}^{3} + {b x}^{2} + c x + d that may have the form

a {(x - h)}^{3} + k, then

c = \frac{b^{2}}{3 a}

Commented by 1549442205PVT last updated on 18/Sep/20

ThankYou . But d is alone! it {isn}^{'} t

arbitrary! so c = \frac{b^{2}}{3 a} {isn}^{'} t enough!

Commented by Dat_Das last updated on 18/Sep/20

I thought about this . After checking that

c = \frac{b^{2}}{3 a} is true and determining h, you set

k = d + {a h}^{3} . This accounts for the only

specification of d and makes it independent

of a, b, c .

Commented by Rasheed.Sindhi last updated on 18/Sep/20

{C o u l d n}^{'} t w e d e t e r m i n e r e l a t i o n

b e t w e e n a, c, d; b, c, d o r a, b, d b y

e x c l u d i n g o n e o f a, b, c ?