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

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(B) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $x$-axis at $(-2,0)$,the point $(-2,0)$ lies on the curve: $0 = a(-2)^3 + b(-2)^2 + c(-2)

Vedclass · Accepted Answer

$\frac{7}{4}$

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(B) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $x$-axis at $(-2,0)$,the point $(-2,0)$ lies on the curve: $0 = a(-2)^3 + b(-2)^2 + c(-2) + 5$ $0 = -8a + 4b - 2c + 5$ $8a - 4b + 2c = 5$ ... $(i)$ Also,the slope of the tangent at $x=-2$ is $0$: $\frac{dy}{dx} = 3ax^2 + 2bx + c$ $3a(-2)^2 + 2b(-2) + c = 0$ $12a - 4b + c = 0$ ... $(ii)$ The curve cuts the $y$-axis at point $Q$. Putting $x=0$,we get $y=5$,so $Q=(0,5)$. The gradient at $Q$ is $3$: $\left(\frac{dy}{dx}\right)_{x=0} = 3 \Rightarrow 3a(0)^2 + 2b(0) + c = 3 \Rightarrow c = 3$. Substitute $c=3$ into $(i)$ and $(ii)$: $(i) \Rightarrow 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1$ ... $(iii)$ $(ii) \Rightarrow 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3$ ... $(iv)$ Subtracting $(iii)$ from $(iv)$: $(12a - 4b) - (8a - 4b) = -3 - (-1) 4a = -2 \Rightarrow a = -\frac{1}{2}$. Substitute $a = -\frac{1}{2}$ into $(iii)$: $8(-\frac{1}{2}) - 4b = -1 -4 - 4b = -1 -4b = 3 \Rightarrow b = -\frac{3}{4}$. Thus,$a+b+c = -\frac{1}{2} - \frac{3}{4} + 3 = \frac{-2-3+12}{4} = \frac{7}{4}$.

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 value of $a+b+c$ is

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