The curve y=ax^3+bx^2+cx+5 touches the X-axis | vedclass

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(C) Given the curve $y=ax^3+bx^2+cx+5$. Since the curve touches the $X$-axis at $(-2,0)$,it passes through $(-2,0)$ and its derivative at $x=-2$ is $0

Vedclass · Accepted Answer

$-\frac{1}{2}, -\frac{3}{4}, 3$

Vedclass · Answer

(C) Given the curve $y=ax^3+bx^2+cx+5$. Since the curve touches the $X$-axis at $(-2,0)$,it passes through $(-2,0)$ and its derivative at $x=-2$ is $0$. Substituting $(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \dots (i)$. Also,the derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$. At $x=-2$,$\frac{dy}{dx} = 0 \Rightarrow 3a(4) + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \dots (ii)$. The curve cuts the $Y$-axis at $Q$. Putting $x=0$ in the curve equation,$y=5$,so $Q$ is $(0,5)$. The gradient at $Q$ is $3$,so $\left(\frac{dy}{dx}\right)_{x=0} = 3 \Rightarrow c = 3$. Substituting $c=3$ into $(i)$ and $(ii)$: $8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \dots (iii)$. $12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \dots (iv)$. Subtracting $(iii)$ from $(iv)$: $4a = -2 \Rightarrow a = -\frac{1}{2}$. Substituting $a = -\frac{1}{2}$ into $(iii)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$. Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $(-2,0)$ and cuts the $Y$-axis at a point $Q$ where its gradient is $3$. Then the values of $a, b, c$ respectively are:

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