# How do you find an equation of the line tangent to #x^2 = -5y# at (5, -5)?

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To find the equation of the line tangent to the curve x^2 = -5y at the point (5, -5), we can use the concept of differentiation.

First, differentiate both sides of the equation with respect to x:

d/dx (x^2) = d/dx (-5y)

2x = -5(dy/dx)

Next, substitute the x-coordinate of the given point (5) into the equation:

2(5) = -5(dy/dx)

10 = -5(dy/dx)

Now, solve for dy/dx:

dy/dx = -10/5

dy/dx = -2

Therefore, the slope of the tangent line at the point (5, -5) is -2.

Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values of the point (5, -5) and the slope (-2):

y - (-5) = -2(x - 5)

y + 5 = -2x + 10

Rearranging the equation to the standard form:

2x + y = 5

Thus, the equation of the line tangent to the curve x^2 = -5y at the point (5, -5) is 2x + y = 5.

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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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