# How do you find the equation of the line tangent to #y=(x^2)e^(x+2)# at x=2?

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To find the equation of the line tangent to the curve y = (x^2)e^(x+2) at x = 2, we need to find the slope of the tangent line at that point and then use the point-slope form of a linear equation.

First, we find the derivative of the given function y = (x^2)e^(x+2) with respect to x. Using the product rule and chain rule, we get:

dy/dx = (2x)e^(x+2) + (x^2)e^(x+2)

Next, we substitute x = 2 into the derivative to find the slope of the tangent line at x = 2:

dy/dx = (2(2))e^(2+2) + (2^2)e^(2+2) = 4e^4 + 4e^4 = 8e^4

So, the slope of the tangent line at x = 2 is 8e^4.

Now, we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope.

Plugging in x1 = 2, y1 = (2^2)e^(2+2) = 16e^4, and m = 8e^4, we have:

y - 16e^4 = 8e^4(x - 2)

This is the equation of the line tangent to y = (x^2)e^(x+2) at x = 2.

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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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