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

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(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point

Vedclass · Accepted Answer

$a=\frac{-1}{2}, b=\frac{-3}{4}, c=3$

Vedclass · Answer

(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point is $0$. Substituting $P(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \quad (1)$. The derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$. At $P(-2,0)$,$\frac{dy}{dx} = 0$: $3a(-2)^2 + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \quad (2)$. The curve cuts the $Y$-axis at $Q(0,k)$. At $Q$,$x=0$ and the gradient is $3$: $\frac{dy}{dx}|_{x=0} = 3(a)(0)^2 + 2(b)(0) + c = 3 \Rightarrow c = 3$. Substituting $c=3$ into $(1)$ and $(2)$: $(1): 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \quad (3)$. $(2): 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \quad (4)$. Subtracting $(3)$ from $(4)$: $(12a - 8a) = -3 - (-1) \Rightarrow 4a = -2 \Rightarrow a = -\frac{1}{2}$. Substituting $a = -\frac{1}{2}$ into $(3)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$. 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 $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where its gradient is $3$. Then:

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