If the tangent to the curve y=x^{3}-x^{2}+x a | vedclass

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(C) The equation of the curve is $y=5x^{2}+2x-25$ at point $P(2, -1)$.First,find the slope of the tangent at $P$ by differentiating: $y' = 10x + 2$.At

Vedclass · Accepted Answer

$195$

Vedclass · Answer

(C) The equation of the curve is $y=5x^{2}+2x-25$ at point $P(2, -1)$.First,find the slope of the tangent at $P$ by differentiating: $y' = 10x + 2$.At $x=2$,the slope $m = 10(2) + 2 = 22$.The equation of the tangent line at $P(2, -1)$ is $y - (-1) = 22(x - 2)$,which simplifies to $y = 22x - 45$.Now,consider the curve $y=x^{3}-x^{2}+x$ at point $(a, b)$.The slope of the tangent at $(a, b)$ is given by $\frac{dy}{dx} = 3x^{2}-2x+1$.At $x=a$,the slope is $3a^{2}-2a+1$.Since this tangent is the same as the one found earlier,its slope must be $22$: $3a^{2}-2a+1 = 22$.This gives the quadratic equation $3a^{2}-2a-21 = 0$.Factoring the quadratic: $(3a+7)(a-3) = 0$,so $a=3$ or $a=-7/3$.For $a=3$,the point $(a, b)$ lies on $y=x^{3}-x^{2}+x$,so $b = 3^{3}-3^{2}+3 = 27-9+3 = 21$.Then $|2a+9b| = |2(3)+9(21)| = |6+189| = 195$.For $a=-7/3$,the tangent line would be parallel but not identical to the line $y=22x-45$,so we reject this value.Thus,the value is $195$.

If the tangent to the curve $y=x^{3}-x^{2}+x$ at the point $(a, b)$ is also tangent to the curve $y=5x^{2}+2x-25$ at the point $(2, -1)$,then $|2a+9b|$ is equal to $........$

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