y=3-4x-x^2

dy/dx=0 when the tangent is horizontal

-2x-4=0

>>>>>>x= -2

plug into original equation

y=f(x)=f(-2)=3+8-4=7

the point where the tangent is horizontal is therefore {-2,7}

Differinate and set to 0.

So, -4-2x=0, or

2x=-4, x=-2.

So,its at (-2,f(-2))=(-2,7).

y=-x^2-4x+3

y’=-2x-4,y’=0,x=-2,y=-4+8+3=7

differetniate it and equate differential to 0

solve and use value of x in the original equation