How do you find the slope of a tangent line to the graph of the function #f(x) = x^2-2x# at the point (a,f(a))?

1 Answer
Oct 31, 2016

# y=(2a-2)x-a^2 #

Explanation:

# f(x)=x^2-2x #

When # x=a => f(a)=a^2-2a #

We get the slope at the tangent at any point by using the derivative,
# f'(x)=2x-2 #
When # x=a => f'(a)=2a-2 #

So, the tangent passes through the point # (a,a^2-2a) # and has slope #m=2a-2#.

Using # y-y_1=m(x-x_1) # the tangent equation is given by:

# y-(a^2-2a)=(2a-2)(x-a) #
# :. y-a^2+2a=2ax-2a^2-2x+2a #
# :. y=2ax-a^2-2x #
# :. y=(2a-2)x-a^2 #