#### Answer

$f(x)= (x+1)^2+2$

#### Work Step by Step

If the vertex of a graph is at $(h, k)$ then the vertex form for the quadratic function is $f(x)=a(x-h)^2+k$.
The vertex is at $(-1,2)$. hence, the tentative equation of the function is:
$f(x)=a[x-(-1)]^2+2\\
f(x)=a(x+1)^2+2.$
$(1,6)$ is on the graph so its $x$ and $y$ coordinates satisfy the equation of the quadratic function.
Hence plugging in the values of $x$ and $y$ of this point into the tentative equation gives
$6=a\cdot(1+1)^2+2\\
6=a(4)+2\\
4=a(4)\\
1=a$
Therefore, $f(x)=(x+1)^2+2$.