The curve y=a x^3+b x^2+c x+5 touches the X-a | vedclass

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(A) Given the curve equation: $y=a x^3+b x^2+c x+5$  ...$(i)$ Since the curve passes through $P(-2,0)$,we substitute $x=-2$ and $y=0$ into equation $(

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$4 a+5$

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(A) Given the curve equation: $y=a x^3+b x^2+c x+5$  ...$(i)$ Since the curve passes through $P(-2,0)$,we substitute $x=-2$ and $y=0$ into equation $(i)$: $0 = a(-2)^3 + b(-2)^2 + c(-2) + 5$ $0 = -8a + 4b - 2c + 5$  ...(ii) Now,find the derivative of the curve: $\frac{dy}{dx} = 3ax^2 + 2bx + c$ Since the curve touches the $X$-axis at $P(-2,0)$,the slope of the tangent at $x=-2$ must be $0$: $\left[\frac{dy}{dx}\right]_{x=-2} = 3a(-2)^2 + 2b(-2) + c = 0$ $12a - 4b + c = 0$ $4b = 12a + c$  ...(iii) Substitute $4b$ from equation (iii) into equation (ii): $-8a + (12a + c) - 2c + 5 = 0$ $4a - c + 5 = 0$ $c = 4a + 5$

The curve $y=a x^3+b x^2+c x+5$ touches the $X$-axis at $P(-2,0)$. Then,$c=$

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