How do you find the tangent line of #f(x) = 3-2x # at x=-1?

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Answer 1 · 2015-11-15T20:41:14

#y=-2x+3#.

At #x=-1, f(-1)=3-2(-1)=5#

So the tangent touches the function at the point #(-1,5)#.

The gradient of the tangent is the derivative of the function.

#therefore f'(x)=-2# and in particular #f'(-1)=-2#.

The tangent is a straight line so has equation #y=mx+c#.

We may substitute the point #(-1,5)# in to obtain

#5=(-2)(-1)+c =>c=3#.

Hence the equation of the required tangent line to the function at the given point is #y=-2x+3#.