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#