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Question Number 44691 by ajfour last updated on 03/Oct/18

Commented by ajfour last updated on 03/Oct/18

$$Findequationofthebiquadratic$$$$curve.$$

Answered by ajfour last updated on 03/Oct/18

$$let\mathit{y}=\mathit{A}(\mathit{x}-a){(\mathit{x}-b)}^2(\mathit{x}-9)+4$$$$andalsolet$$$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=4\mathit{A}(\mathit{x}-2)(\mathit{x}-\mathit{b})(\mathit{x}-\mathit{c})$$$$atx=c,y=0;hence$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\mathit{A}(\mathit{c}-9){(\mathit{c}-\mathit{b})}^2(\mathit{c}-\mathit{a})=-4..\left(i\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$atx=0,y=-2,so$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$3{\mathit{A}\mathit{a}\mathit{b}}^2=-2....\left(ii\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\frac{dy}{dx}=\frac{d}{dx}\left(y\right)implies$$$$4A(x-b)(x-2)(x-c)=$$$$=A[{(x-b)}^2(x-9)+2(x-b)(x-a)(x-9)$$$$+(x-a){(x-b)}^2]$$$$\Rightarrow$$$$4(x-2)(x-c)=4x^2+(-b-9-2a-18$$$$-a-b)x$$$$+(9b+18a+ab)$$$$comparingcoefficientsof{x}^{1}on$$$$bothsides:$$$$\Rightarrow 8+4c=3a+2b+27$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow 3\mathit{a}+2\mathit{b}+19=4\mathit{c}....\left(iii\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$comparingconstanttermsof$$$$\frac{dy}{dx}\mathrm{\&}\frac{d}{dx}\left(y\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\mathit{a}\mathit{b}+18\mathit{a}+9\mathit{b}=8\mathit{c}....\left(iv\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$...$$$$ultimatelyeliminating\mathit{A},\mathit{b},\mathit{c}$$$$amongthefourequations:$$$$24\times 256\mathit{a}{(19-6\mathit{a})}^2{(5+\mathit{a})}^2$$$$=(3{\mathit{a}}^2-26\mathit{a}-9){(3{\mathit{a}}^2+58\mathit{a}+19)}^2$$$$\times (171-{\mathit{a}}^2-10\mathit{a})$$$$\Rightarrow \mathit{a}=-0.0092$$$$\mathit{b}=7.63613$$$$\mathit{c}=8.5612$$$$\mathit{A}=1.242723$$$$soeq.is$$$$y=1.242723(x+0.0092)$$$$\times {(x-7.63613)}^2(x-9)+4.$$$$Pleasecheck..$$

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