The curve $y\left(x\right)=a{x}^3+b{x}^2+cx+5$ touches the x-axis at the point P(–2, 0) and cuts the y-axis at the point Q, where y' is equal to 3. Then the local maximum value of y(x) is :

$y\left(x\right)=a{x}^3+b{x}^2+cx+5$ is passing through (-2,0) then $8a-4b+2c=5$ .....(1)

$y^{\mathrm{\prime }}\left(x\right)=3a{x}^2+2bx+c$touches x-axis at (-2,0)

$12a-4b+c=0$.....(2)

again, for $x=0,y^{\mathrm{\prime }}\left(x\right)=3$

$c=3$......(3)

Solving eq. (1), (2) & (3) $a=-\frac{1}{2},b=-\frac{3}{4}$

$y^{\mathrm{\prime }}\left(x\right)=-\frac{3}{2}{x}^2-\frac{3}{2}x+3$

y(x) has local maxima at x = 1

$y\left(1\right)=\frac{27}{4}$