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

Question

Cameron Alexander · Accepted Answer

#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!

Emma Aldridge · Answer

undefined To find the equation of the tangent line to the graph of f(x) = e^(-2x+x^2) through the point (2,1), we need to find the slope of the tangent line at that point.

First, we find the derivative of f(x) with respect to x.

f'(x) = (-2 + 2x)e^(-2x+x^2)

Next, we substitute x = 2 into f'(x) to find the slope at x = 2.

f'(2) = (-2 + 2(2))e^(-2(2)+2^2) = 0

Since the slope is 0, the tangent line is horizontal.

The equation of a horizontal line passing through the point (2,1) is y = 1.

Therefore, the equation of the tangent line to the graph of f(x) = e^(-2x+x^2) through the point (2,1) is y = 1.

Isabella Ackerman · Answer

undefined To find the equation of the tangent line to the graph of $ f(x) = e^{-2x + x^2} $ through the point $(2, 1)$, first, find the derivative of $ f(x) $ using the power rule and the chain rule. Then, evaluate the derivative at $ x = 2 $ to find the slope of the tangent line. Finally, use the point-slope form of a linear equation to write the equation of the tangent line.