The curve y(x) = a x ^ 3 + b x ^ 2 + cx + 5 touches the x -axis at...

Given that $y\left(x\right) = a{x}^{3} + b{x}^{2} + cx + 5$ pass through \left(- 2, 0\right)$</p><p>so$8a - 4b + 2c = 5 . . . \left(i\right)

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

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

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

Also given, for $x = 0, {y}^{'}( x ) = 3$

$c = 3 . . . \left(iii\right)</p><p>Solving eq.$\left(i\right), \left(ii\right)\text{and }\left(iii\right), we get, </p><p>a = - $\dfrac{1}{2}$, b = - $\dfrac{3}{4}</p><p>For local maxima${y}^{'}\left(x\right)$= -$\dfrac{3}{2}${x}^{2} -$\dfrac{3}{2}$x + 3 = 0$

$⇒ {x}^{2} + x - 2 = 0 ⇒ \left(x - 1\right)$\left(x + 2\right)$= 0$

⇒ x = 1 and ${y}^{"}\left(1\right)$ < 0$</p><p>So$y\left(x\right)$has local maxima at$x = 1$</p><p>Hence,$y\left(1\right) = \dfrac{27}{4}