How do you find the equation of the tangent line to the graph #f(x)=e^(-2x+x^2)# through point (2,1)?

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Answer 1 · 2016-12-14T02:00:23

#y = 2x - 3#

Use the chain rule to differentiate.

Let #y= e^u# and #u = x^2 - 2x#.

Then #dy/(du)= e^u# and #(du)/dx= 2x - 2#.

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx =e^u xx 2x - 2#

#dy/dx = (2x - 2)e^(x^2 - 2x)#

The slope of the tangent is given by evaluating your point #x = a# into the derivative.

#m_"tangent" = (2(2) - 2)e^(2^2 - 4)#

#m_"tangent" = 2(e^0)#

#m_"tangent" = 2#

The equation is therefore:

#y- y_1 = m(x - x_1)#

#y - 1 = 2(x - 2)#

#y - 1 = 2x - 4#

#y = 2x - 3#

Hopefully this helps!