Given that y(x)=ax3+bx2+cx+5 pass through (−2,0)
so 8a−4b+2c=5...(i)
Since the curve touches x-axis at (−2,0), so its slope would be 0
i.e y′(−2)=0⇒(3ax2+2bx+c)x=−2=0
12a−4b+c=0...(ii)
Also given, for x=0,y′(x)=3
c=3...(iii)
Solving eq. (i),(ii)&(iii), we get,
a=−21,b=−43
For local maxima y′(x)=−23x2−23x+3=0
⇒x2+x−2=0⇒(x−1)(x+2)=0
⇒x=1 and y"(1)<0
So y(x) has local maxima at x=1
Hence, y(1)=427