How do you find the equation of the tangent and normal line to the curve $y=x^2-3$ at (2,1)?

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Answer 1 · 2016-10-06T02:57:08

$y = 4 x - 7$

The slope, m, of the tangent line at any given point is the first derivative, y'(x), evaluated at the x coordinate of the given point.

$y'(x) = 2x$

$y'(2) = 2(2) = 4$

$m = 4$

Using the $y = m(x) + b$ form, we substitute, 1 for y, 2 for x, 4 for m, and then solve for b:

$1 = 4(2) + b$

$b = -7$

The equation of the tangent line is:

$y = 4 x - 7$

How do you find the equation of the tangent and normal line to the curve y=x^2-3y=x^2-3 at (2,1)?