How do you find the slope and equation of the tangent line to the curve $y=9-2x^2$ at the point (2,1)?

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Answer 1 · 2015-08-31T01:14:20

$y = - 8x + 17$

Given -

$y = 9 - 2x^2$

Its first derivative gives the slope of the curve st any given point.

$dy/dx = - 4x$

The tangent passes through the point (2, 1)

At x = 2 ; the slope of the given curve is

$m = -4 * (2) = -8$

Slope of the tangent is the same as slope of the curve at point (2,1)

Equation of the line -

$y - y_1 = m (x - x_1)$

$y - 1 = - 8 (x - 2)$

$y -1 = - 8x +16$

$y = - 8x + 17$

How do you find the slope and equation of the tangent line to the curve y=9-2x^2y=9-2x^2 at the point (2,1)?