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 :)