The curve y(x)=ax^3+bx^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…

Given that $y(x)=a{x}^{3}+b{x}^{2}+cx+5$y(x)=ax3+bx2+cx+5 pass through $(-2,0)$(−2,0)

so $8a-4b+2c=5...(i)$8a−4b+2c=5...(i)

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

i.e ${y}^{'}(-2)=0\Rightarrow {(3a{x}^{2}+2bx+c)}_{x=-2}=0$y′(−2)=0⇒(3ax2+2bx+c)x=−2​=0

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

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

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

Solving eq. (i),(ii)&(iii), we get,

$$a=-\frac{1}{2},b=-\frac{3}{4}$$a=−21​,b=−43​

For local maxima ${y}^{'}(x)=-\frac{3}{2}{x}^{2}-\frac{3}{2}x+3=0$y′(x)=−23​x2−23​x+3=0

$$\Rightarrow {x}^{2}+x-2=0\Rightarrow (x-1)(x+2)=0$$⇒x2+x−2=0⇒(x−1)(x+2)=0

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

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

Hence, $y(1)=\frac{27}{4}$y(1)=427​