What is the equation of the tangent line of #f(x)=-2x^2+5x+2 # at #x=-1#?

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Answer 1 · 2015-12-08T11:27:35

#y=9x+4#

To find a line, we must know its slope and one of its points.

The slope is given by the derivative evaluated at the requested point, so we have

#f'(x)=-4x+5 \implies f'(-1)=4+5=9#

The point is of course the point on the graph #(-1, f(-1))#, which is (-1, -5)#

Once the slope and a point are known, the equation of the line is given by

#y-y_0=m(x-x_0)#

#y+5 = 9(x+1)#

#y=9x+9-5=9x+4#