# How do you use the vertical line test to show y=3- sqrt(x+2) is a function?

Jul 9, 2015

There are no $\left(x , {y}_{1}\right) , \left(x , {y}_{2}\right)$ on the graph$\left(f\right)$ such that ${y}_{1} \setminus \ne {y}_{2}$

#### Explanation:

We want to show that if $\left(x , {y}_{1}\right) \setminus \in \setminus \Gamma \left(f\right)$ and $\left(x , {y}_{2}\right) \setminus \in \setminus \Gamma \left(f\right)$ then ${y}_{1} = {y}_{2}$

${y}_{1} = 3 - \sqrt{x + 2}$

${y}_{2} = 3 - \sqrt{x + 2}$

${y}_{1} - {y}_{2} = 3 - 3 = 0$

${y}_{1} = {y}_{2}$