How do you make a curve y=ax^3 +bx^2 +cx+d with a critical point of (1,0) and (−2,27) ?


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Question Number 187053 by horsebrand11 last updated on 13/Feb/23

$H o w d o y o u m a k e a c u r v e$ $y = {a x}^{3} + {b x}^{2} + c x + d w i t h a c r i t i c a l$ $p o i n t o f (1, 0) a n d (- 2, 27) ?$
	Answered by cortano12 last updated on 13/Feb/23
	$⇒3ax^2 +2bx+c = k(x−1)(x+2) ⇒3ax^2 +2bx+c = kx^2 +kx−2k ⇒ { ((k=3a=2b ; b=(3/2)a)),((c=−2k=−6a)) :} ∵ y=ax^3 +(3/2)ax^2 −6ax+d ⇒−(7/2)a+d=0⇒d=(7/2)a ⇒10a+d=27 ⇒((27)/2)a=27 ; a=2 { ((d=7)),((c=−12)),((b=3)) :} ⇒y=2x^3 +3x^2 −12x+7$
	$\Rightarrow 3 {a x}^{2} + 2 b x + c = k (x - 1) (x + 2)$ $\Rightarrow 3 {a x}^{2} + 2 b x + c = {k x}^{2} + k x - 2 k$ $\Rightarrow {\begin{cases} k = 3 a = 2 b; b = \frac{3}{2} a \\ c = - 2 k = - 6 a \end{cases}$ $∵ y = {a x}^{3} + \frac{3}{2} {a x}^{2} - 6 a x + d$ $\Rightarrow - \frac{7}{2} a + d = 0 \Rightarrow d = \frac{7}{2} a$ $\Rightarrow 10 a + d = 27 \Rightarrow \frac{27}{2} a = 27; a = 2$ ${\begin{cases} d = 7 \\ c = - 12 \\ b = 3 \end{cases} \Rightarrow y = 2 x^{3} + 3 x^{2} - 12 x + 7$
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