How do you find the equation of a line tangent to the function #y=x-2x^2+3# at x=2?

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Answer 1 · 2017-01-11T00:49:57

# y = -7x+11 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

so If # y = x-2x^2+3 # then differentiating wrt #x# gives us:

# dy/dx = 1-4x #

When #x=2 => y=2-8+3=-3# (so #(2,-3)# lies on the curve)
and # dy/dx = 1-8=-7#

So the tangent passes through #(2,-3)# and has gradient #-7#, so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-(-3) = -7(x-2) #
# :. y+3 = -7x+14#
# :. \ \ \ \ \ \ \ y = -7x+11 #

We can confirm this solution is correct graphically:
enter image source here