So A=2, φ=-π/6.

f(0)=2sin(-π/6)=- 1

Question 2

If the tangent equation crosses a point, find the value of f( 1), and the slope is the derivative value of the function, and find the value of f' (1).

f( 1)=e( 1^2+a-a)=e

f'(x)=e^x(x^2+ax-a)+e^x(2*x+a)=e^xx^2+(2+a)x-a

f '( 1)= e( 1+2+a)=(3+a)e

The equation is y-e=(3+a)e(x- 1).