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First vector orthogonal<span> to ⟨−</span>3<span>,4</span>
We have been given a quadratic function
and we need to restrict the domain such that it becomes a one to one function.
We know that vertex of this quadratic function occurs at (5,2).
Further, we know that range of this function is
.
If we restrict the domain of this function to either
or
, it will become one to one function.
Let us know find its inverse.

Upon interchanging x and y, we get:

Let us now solve this function for y.

Hence, the inverse function would be
if we restrict the domain of original function to
and the inverse function would be
if we restrict the domain to
.