# How do you find the equation of the tangent line to the curve #f(x) = x^2 + 5x# at x = 4?

Like this:

graph{(13x-16-y)(x^2+5x-y)=0 [-64.1, 64, -32, 32.1]}

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To find the equation of the tangent line to the curve f(x) = x^2 + 5x at x = 4, we need to find the slope of the tangent line and a point on the line.

First, find the derivative of f(x) with respect to x, which gives us f'(x) = 2x + 5.

Next, substitute x = 4 into f'(x) to find the slope of the tangent line at x = 4. f'(4) = 2(4) + 5 = 13.

Now, we have the slope of the tangent line, which is 13.

To find a point on the line, substitute x = 4 into the original function f(x). f(4) = (4)^2 + 5(4) = 36.

Therefore, the point on the tangent line is (4, 36).

Using the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values we found to get the equation of the tangent line.

So, the equation of the tangent line to the curve f(x) = x^2 + 5x at x = 4 is y = 13x - 16.

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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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