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

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(A) The equation of the curve is $y=ax^3+bx^2+cx+5$ ...$(i)$Since the curve touches the $X$-axis at $P(-2,0)$,it passes through $P(-2,0)$,so $-8a+4b-2

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

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(A) The equation of the curve is $y=ax^3+bx^2+cx+5$ ...$(i)$Since the curve touches the $X$-axis at $P(-2,0)$,it passes through $P(-2,0)$,so $-8a+4b-2c+5=0$ ...(ii)Also,since it touches the $X$-axis at $x=-2$,the derivative $\frac{dy}{dx}$ at $x=-2$ is $0$.$\frac{dy}{dx} = 3ax^2+2bx+c$. At $x=-2$,$12a-4b+c=0$ ...(iii)The curve cuts the $Y$-axis at $Q(0,5)$ where the gradient is $3$,so $\left.\frac{dy}{dx}\right|_{x=0} = 3$,which gives $c=3$.Substituting $c=3$ into (ii) and (iii):$-8a+4b-6+5=0 \Rightarrow -8a+4b=1$ ...(iv)$12a-4b+3=0 \Rightarrow 12a-4b=-3$ ...$(v)$Adding (iv) and $(v)$: $4a = -2 \Rightarrow a=-\frac{1}{2}$.Substituting $a=-\frac{1}{2}$ into (iv): $-8(-\frac{1}{2})+4b=1 \Rightarrow 4+4b=1 \Rightarrow 4b=-3 \Rightarrow b=-\frac{3}{4}$.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 the gradient is $3$. Then the values of $a, b, c$ are

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