Assuming the question is asking about $y=2+(x-3)^2$
Answer:
$(3, \infty)$
Step-by-step explanation:
We see that the function is quadratic, and it is already in vertex form. This is good, as it saves a step. Because anything squared is always positive or $0$, when we minimize this function we want to have the thing inside the square be $0$. This means $x-3=0$, so $x=3$.
We can plug in for $x$ to find the vertex.
$y=2+(3-3)^2$
$y=2+0^2$
$y=2+0$
$y=2$.
This means that $(3, 2)$ is the vertex of the parabola(a quadratic equation in $x$).
We see that the coefficient of the highest order term(the $x^2$) will be positive, so the parabola faces upward. This means it will always be increasing to the right of the vertex, and decreasing to the left of the vertex.
So, the interval where $y=2+(x-3)^2$ is increasing is the interval to the right of the vertex, which is $(3, \infty)$
