How do you find the equation of a line tangent to the function #y=6-x^2# at (2,2)?

1 Answer
Jan 30, 2017

The equation of the line is #y = -4x + 10#

Explanation:

The function #6-x^2# is a polynomial, which is definitely differentiable, with #y' = -2x#. That #-2x# is also the slope of the line tangent to the graph of #y# at any given point #x#.

Since we are trying to find the tangent at #(2,2)#, we know that the slope will be #-2 * 2 = -4#. So, the equation of the line is:

#y = -4x + a#, where #a# is the point at which the line intersects the #y# axis. To find #a#, simply plug in the #x# and #y# coordinates of the point:

#2 = -8 + a => a = 10#. The final equation is:

#y = -4x + 10#.