# How do you find the equation of the line tangent to #y= -x^3+6x^2-5x# at (1,0)?

eqn of tangent

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To find the equation of the line tangent to the curve y = -x^3 + 6x^2 - 5x at the point (1,0), we need to find the slope of the tangent line at that point.

First, we find the derivative of the given function y = -x^3 + 6x^2 - 5x.

The derivative of y with respect to x is dy/dx = -3x^2 + 12x - 5.

Next, we substitute x = 1 into the derivative to find the slope at the point (1,0).

dy/dx = -3(1)^2 + 12(1) - 5 = 4.

Therefore, the slope of the tangent line at (1,0) is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point 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 - 0 = 4(x - 1).

Simplifying, we get y = 4x - 4.

Thus, the equation of the line tangent to y = -x^3 + 6x^2 - 5x at (1,0) is y = 4x - 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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