# How do you find the equation of the line tangent to the graph of #y = e^(-x^2)# at the point (2, 1/e^4)?

- find derivative of function
- sub in x value of point to find gradient of tangent
- put gradient into y=(gradient)x+c
- Sub in point and solve for c
- you have found the equation of the tangent.

so now we have to sub in the point and solve for c:

hope that helped.

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To find the equation of the line tangent to the graph of y = e^(-x^2) at the point (2, 1/e^4), we need to find the derivative of the function y = e^(-x^2) and evaluate it at x = 2.

The derivative of y = e^(-x^2) can be found using the chain rule. The derivative of e^(-x^2) with respect to x is -2x * e^(-x^2).

Evaluating the derivative at x = 2, we have -2(2) * e^(-2^2) = -4 * e^(-4).

So, the slope of the tangent line at x = 2 is -4 * e^(-4).

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point (2, 1/e^4) and m is the slope, we can substitute the values to find the equation of the tangent line.

Plugging in the values, we have y - 1/e^4 = -4 * e^(-4)(x - 2).

Simplifying further, we get y - 1/e^4 = -4e^(-4)x + 8e^(-4).

Rearranging the equation, we have y = -4e^(-4)x + 8e^(-4) + 1/e^4.

Therefore, the equation of the line tangent to the graph of y = e^(-x^2) at the point (2, 1/e^4) is y = -4e^(-4)x + 8e^(-4) + 1/e^4.

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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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