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

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Answer 1 · 2016-11-01T23:18:09

Tangent $y = x-1$
Normal $y=-x+1$

We have $y=x^2-x$

Differentiating wrt $x$ we get:
$dy/dx=2x-1$

When $x=1 => y=1-1=0$
And, $x=1 => dy/dx = 2-1=1$

So the tangent passes through the point (1,0) and has gradient $m_T=1$ , Using $y-y_1=m(x-x_1)$ , so the equation of the tangent is:

$y-0 = (1)(x-1)$
$:. y = x-1$

The normal is perpendicular to the tangent,so the product of their gradients is $-1$ , so the gradient of the normal is $m_N =-1/1 = -1$

So the normal passes through the point (1,0) and has gradient $m_N =-1$ , so the equation of the normal is:

$y-0=(-1)(x-1)$
$y=-x+1$

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