What is the equation of the tangent line of #f(x)=x^2 - 2x # at #x=5#?

1 Answer
Alicia
May 8, 2017

#2y = x + 25#

Explanation:

So at that point on the curve, as it is a tangent the gradient will be the same.
The gradient is calculated by differentiating.
#dy/dx x^2 - 2x dx = 1/2x - 2#
We can then substitute #x# in to find the gradient at #x=5#.
Therefore, #1/2(5) - 2 = 1/2#.
This is our gradient.
Then we need to find #y# at the point #x = 5#.
Substitute into #x^2 - 2x# to get #(5)^2 -2(5) = 15#.
Now we can use the equation of #y - y_1 = m(x - x_1)#.
Therefore:
#y - 15 = 1/2(x - 5)#
Expand and rearrange:
#y - 15 = 1/2x - 5/2#
#y = 1/2x + 25/2#
#2y = x + 25#.

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