How do you find the slope of a tangent line to the graph of the function # f(x)= (2x^2+5x-4)/x^2# at x=3?

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Answer 1 · 2017-01-12T22:09:33

# y = -7/27x+4 #

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 #f(x) = (2x^2+5x-4)/x^2 =2+5/x-4/x^2# then differentiating wrt #x# gives us:

# f'(x) = -5/x^2+8/x^3 #

When #x=3 => f(3)=(18+15-4)/9=29/9#
and # f'(3) = -5/9+8/27=-7/27#

So the tangent passes through #(3,29/9)# and has gradient #-7/27#, so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-(29/9) = -7/27(x-3) #
# :. y-29/9 = -7/27x+7/9 #
# :. \ \ \ \ \ \ \ y = -7/27x+4 #

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