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

Use the chain rule to differentiate.

The equation is therefore:

Hopefully this helps!

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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.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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