The curve y x = a x 3 + b x 2 + c x + 5 touches the x -axis at the point P - 2 , 0 and cuts the y -axis at…

Given that $y\left(x\right)=a{x}^3+b{x}^2+cx+5$ pass through $\left(-2,0\right)$ 

so $8a-4b+2c=5 ...\left(i\right)$

Since the curve touches $x$-axis at $\left(-2,0\right)$, so its slope would be $0$

i.e $y^{\text{'}}\left(-2\right)=0\Rightarrow {\left(3a{x}^2+2bx+c\right)}_{x=-2}=0$

$12a-4b+c=0 ...\left(ii\right)$

Also given, for $x=0, y^{\text{'}}\mathit{\left(}\mathit{x}\mathit{\right)}=3$

$c=3 ...\left(iii\right)$

Solving eq. $\left(i\right), \left(ii\right) \& \left(iii\right)$, we get, 

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

For local maxima $y^{\text{'}}\left(x\right)=-\frac{3}{2}{x}^2-\frac{3}{2}x+3=0$

$\Rightarrow x^2+x-2=0\Rightarrow \left(x-1\right)\left(x+2\right)=0$

$\Rightarrow x=1$ and $y^{"}\left(1\right)<0$

So $y\left(x\right)$ has local maxima at $x=1$

Hence, $y\left(1\right)=\frac{27}{4}$