Precalculus (10th Edition)

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