we are given the following standard form of a function:
![f(x)=\sqrt[]{x-h}+k](https://img.qammunity.org/2023/formulas/mathematics/college/pm28npxctbhqc7wo2iqifj9i29nurdb4vf.png)
We are also given the following function:
![f(x)=\sqrt[]{x+2}+1](https://img.qammunity.org/2023/formulas/mathematics/college/xeu81kmpmkue2ejvvx5p0jhu4u24qrj8o6.png)
since "x - h" is the term under the radical, we have:
![x-h=x+2](https://img.qammunity.org/2023/formulas/mathematics/college/4lm35aeqd4u88yub4en64nlb18tpcd7mg4.png)
setting this value to zero, we get:
![x+2=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/cxsi5axofj7xqk6n84auu1u1x233fshjvi.png)
To solve for "x" we will subtract 2 to both sides:
![\begin{gathered} x+2-2=-2 \\ x=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1swnchwp2jlpuguksbmsheu9o2n1wukw3i.png)
"k" is the value we add to the radical, therefore:
![k=1](https://img.qammunity.org/2023/formulas/mathematics/high-school/lmua7ym7ajkn75clv8n2iars4aey6fnqkp.png)
The starting point is:
![(h,k)=(-2,1)](https://img.qammunity.org/2023/formulas/mathematics/college/acyedw7b0gqpa7o4jcz31urnll5bfucmfz.png)
That is, up one, right two.