# How do you find the equation of a line tangent to the function #y=x^3-6x+1# at x=2?

We have

# y =x^3-6x+1 #

The gradient of the tangent at any particular point is given by the derivative: So differntiating wrt

# dy/dx=3x^2-6 #

When

And,

So the tangent passes through

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To find the equation of a line tangent to a function at a specific point, you can follow these steps:

- Find the derivative of the function.
- Substitute the given x-coordinate into the derivative to find the slope of the tangent line.
- Use the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope, to write the equation of the tangent line.

For the function y = x^3 - 6x + 1, the derivative is dy/dx = 3x^2 - 6.

Substituting x = 2 into the derivative, we get dy/dx = 3(2)^2 - 6 = 6.

The slope of the tangent line at x = 2 is 6.

Using the point-slope form with the point (2, f(2)), where f(2) is the value of the function at x = 2, we can write the equation of the tangent line.

Substituting x = 2 into the original function, we get f(2) = (2)^3 - 6(2) + 1 = 1.

Therefore, the point on the tangent line is (2, 1).

Using the point-slope form, the equation of the tangent line is y - 1 = 6(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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