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

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

$$\mathrm{For}\:\mathrm{a}\:\mathrm{cubic}\:\mathrm{function}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}: \\ $$$${f}\left({x}\right)\:=\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$$\mathrm{What}\:\mathrm{must}\:\mathrm{be}\:\mathrm{true}\:\mathrm{of}\:{a},\:{b},\:{c},\:\mathrm{and}\:{d}\:\mathrm{in} \\ $$$$\mathrm{order}\:\mathrm{for}\:\mathrm{the}\:\mathrm{function}\:\mathrm{to}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{converted}\:\mathrm{to}\:\mathrm{the}\:\mathrm{form}: \\ $$$${f}\left({x}\right)\:=\:{a}\left({x}−{h}\right)^{\mathrm{3}} +{k} \\ $$

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

We need must find a,b,c,d such that ax^3 +bx^2 +cx+d=a(x−h)^3 +k ⇔ ax^3 +bx^2 +cx+d≡ax^3 −3ahx^2 +3ah^2 x−ah^3 +k ⇔ { ((b=−3ah)),((c=3ah^2 )),((d=−ah^3 +k)) :} where a is arbitrary;h,k given

$$\mathrm{We}\:\mathrm{need}\:\mathrm{must}\:\mathrm{find}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}={a}\left({x}−{h}\right)^{\mathrm{3}} +{k} \\ $$$$\Leftrightarrow\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\equiv\mathrm{ax}^{\mathrm{3}} −\mathrm{3ahx}^{\mathrm{2}} +\mathrm{3ah}^{\mathrm{2}} \mathrm{x}−\mathrm{ah}^{\mathrm{3}} +\mathrm{k} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{b}=−\mathrm{3ah}}\\{\mathrm{c}=\mathrm{3ah}^{\mathrm{2}} }\\{\mathrm{d}=−\mathrm{ah}^{\mathrm{3}} +\mathrm{k}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{is}\:\mathrm{arbitrary};\mathrm{h},\mathrm{k}\:\mathrm{given} \\ $$$$ \\ $$

Commented by Dat_Das last updated on 18/Sep/20

This is true, however you may relate a, b, c directly. Take b=−3ah (b/(−3a))=h (b^2 /(9a^2 ))=h^2 (b^2 /(3a))=3ah^2 Substituting c for 3ah^2 we get (b^2 /(3a))=c Therefore for any function ax^3 +bx^2 +cx+d that may have the form a(x−h)^3 +k, then c=(b^2 /(3a))

$$\mathrm{This}\:\mathrm{is}\:\mathrm{true},\:\mathrm{however}\:\mathrm{you}\:\mathrm{may}\:\mathrm{relate} \\ $$$${a},\:{b},\:{c}\:\mathrm{directly}.\:\mathrm{Take} \\ $$$${b}=−\mathrm{3}{ah} \\ $$$$\frac{{b}}{−\mathrm{3}{a}}={h} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{9}{a}^{\mathrm{2}} }={h}^{\mathrm{2}} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}=\mathrm{3}{ah}^{\mathrm{2}} \\ $$$$\mathrm{Substituting}\:{c}\:\mathrm{for}\:\mathrm{3}{ah}^{\mathrm{2}} \:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}={c} \\ $$$$\mathrm{Therefore}\:\mathrm{for}\:\mathrm{any}\:\mathrm{function} \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\:\mathrm{that}\:\mathrm{may}\:\mathrm{have}\:\mathrm{the}\:\mathrm{form} \\ $$$${a}\left({x}−{h}\right)^{\mathrm{3}} +{k},\:\mathrm{then} \\ $$$${c}=\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$

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

ThankYou.But d is alone!it isn′t arbitrary!so c=(b^2 /(3a)) isn′t enough!

$$\mathrm{ThankYou}.\mathrm{But}\:\mathrm{d}\:\mathrm{is}\:\mathrm{alone}!\mathrm{it}\:\mathrm{isn}'\mathrm{t} \\ $$$$\mathrm{arbitrary}!\mathrm{so}\:\mathrm{c}=\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{3a}}\:\mathrm{isn}'\mathrm{t}\:\mathrm{enough}! \\ $$

Commented by Dat_Das last updated on 18/Sep/20

I thought about this. After checking that c=(b^2 /(3a)) is true and determining h, you set k=d+ah^(3 ) . This accounts for the only specification of d and makes it independent of a, b, c.

$$\mathrm{I}\:\mathrm{thought}\:\mathrm{about}\:\mathrm{this}.\:\mathrm{After}\:\mathrm{checking}\:\mathrm{that} \\ $$$${c}=\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}}\:\mathrm{is}\:\mathrm{true}\:\mathrm{and}\:\mathrm{determining}\:{h},\:\mathrm{you}\:\mathrm{set} \\ $$$${k}={d}+{ah}^{\mathrm{3}\:} .\:\mathrm{This}\:\mathrm{accounts}\:\mathrm{for}\:\mathrm{the}\:\mathrm{only} \\ $$$$\mathrm{specification}\:\mathrm{of}\:{d}\:\mathrm{and}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{independent} \\ $$$$\mathrm{of}\:{a},\:{b},\:{c}. \\ $$

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

Couldn′t we determine relation between a,c,d ; b,c,d or a,b,d by excluding one of a,b,c ?

$${Couldn}'{t}\:{we}\:{determine}\:{relation} \\ $$$${between}\:{a},{c},{d}\:;\:{b},{c},{d}\:{or}\:{a},{b},{d}\:{by} \\ $$$${excluding}\:{one}\:{of}\:{a},{b},{c}\:? \\ $$

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