# How do you find the slope of the tangent line to the graph of the function f(x)=x^2-4 at (1,-3)?

May 12, 2017

See explanation.

#### Explanation:

The slope of a tangent line at $\left({x}_{0} , f \left({x}_{0}\right)\right)$ is equal to the value of the first derivative at ${x}_{0}$: ${f}^{'} \left({x}_{0}\right)$.

Here we have:

$f \left(x\right) = {x}^{2} - 4$

${f}^{'} \left(x\right) = 2 x$

${f}^{'} \left(1\right) = 2$

Answer The slope of the tangent line at ${x}_{0} = 1$ is $2$.