How do you write the slope of the line tangent to #g(x)=x^2-4# at the point (1,-3)?

1 Answer
Nov 11, 2017

2

Explanation:

First, we know that #dy/dx# is the slope of the curve at any point.
So, we can find #g'(x)# at the beginning.

#g'(x)=d/dx(x^2-4)=d/dx(x^2)-d/dx(4)=2x-0=2x#

Then, at the point #(1,-3)# , the slope of the line tangent to #g(x)# is just #g'(1)#, by replacing the value of #x#.

Therefore, the slope of tangent at point #(1,-3)#
#=g'(1)=2(1)=2#

Here is the answer. Hope it can help you :)