# How do you find the equation of the tangent and normal line to the curve #y=x^2-x# at x=1?

Tangent

Normal

We have

Differentiating wrt

When

And,

So the tangent passes through the point (1,0) and has gradient

The normal is perpendicular to the tangent,so the product of their gradients is

So the normal passes through the point (1,0) and has gradient

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To find the equation of the tangent line to the curve (y = x^2 - x) at (x = 1), first find the derivative of the function, which represents the slope of the tangent line. Then, evaluate the derivative at (x = 1) to find the slope of the tangent line at that point. After that, use the point-slope form of the equation of a line, (y - y_1 = m(x - x_1)), where (m) is the slope of the tangent line and ((x_1, y_1)) is the point of tangency, to write the equation of the tangent line.

To find the equation of the normal line, find the negative reciprocal of the slope of the tangent line (since the tangent and normal lines are perpendicular), and use the same point of tangency to write the equation of the normal line using the point-slope form.

The process can be summarized as follows:

- Find the derivative of (y = x^2 - x).
- Evaluate the derivative at (x = 1) to find the slope of the tangent line.
- Use the point-slope form to write the equation of the tangent line.
- Find the negative reciprocal of the slope of the tangent line to get the slope of the normal line.
- Use the point-slope form to write the equation of the normal line.

Once you follow these steps, you'll have the equations of both the tangent and normal lines to the curve (y = x^2 - x) at (x = 1).

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