At what point is the tangent to f(x)=3-4x-x^2 horizontal?

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

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