=3ax2+2bx+c‌‌...(ii) Touches x-axis at P(−2,0) .⇒y|x=−2=0⇒−8a+4b−2c+5=0‌‌......‌ (iii) ‌ Touches x-axis at P(−2,0) also implies .‌
dy
dx
|x=−2=0⇒12a−4b+c=0‌‌......(iv) y=f(x) cuts y-axis at (0,5) Given, .‌
dy
dx
|x=0=c=3 From (iii), (iv) and (v) a=−‌
1
2
,b=−‌
3
4
,c=3 ⇒f(x)=‌
−x2
2
−‌
3
4
x2+3x+5 f′(x)=‌
−3
2
x2−‌
3
2
x+3 =‌
−3
2
(x+2)(x−1) f′(x)=0 at x=−2 and x=1 By first derivative test x=1 in point of local maximum Hence local maximum value of f(x) is f(1) i.e., ‌