Question

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

Answer 1

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 :)