If a and b are constants, and the graph of g(x) = (x − a)2 + b has its minimum at g(6), which of the following pairs of points could also be on the graph of g(x) in the xy-plane?

The function is given as $g(x) = (x-a)^2 + b$. This is the vertex form of a parabola, $y = (x-h)^2 + k$, where $(h, k)$ is the vertex.
For $g(x)$, the vertex is $(a, b)$. Since the coefficient of $(x-a)^2$ is positive (it's 1), the parabola opens upwards, meaning the vertex is the minimum point.

The problem states that the graph has its minimum at $g(6)$. This means the x-coordinate of the vertex is 6.
Therefore, $a = 6$.
The function can now be written as $g(x) = (x-6)^2 + b$.

A parabola is symmetric about its axis of symmetry, which is a vertical line passing through its vertex. For $g(x) = (x-6)^2 + b$, the axis of symmetry is $x=6$.
This means that if two distinct points on the parabola have the same y-coordinate, their x-coordinates must be equidistant from the axis of symmetry $x=6$.
If $(x_1, y)$ and $(x_2, y)$ are two points on the parabola, then $y = (x_1-6)^2 + b$ and $y = (x_2-6)^2 + b$.
This implies $(x_1-6)^2 = (x_2-6)^2$.
Taking the square root of both sides gives $x_1-6 = \pm (x_2-6)$.
Since the points are distinct, $x_1 \neq x_2$, so we must have $x_1-6 = -(x_2-6)$.
$x_1-6 = -x_2+6$
$x_1+x_2 = 12$.
So, for any two distinct points with the same y-coordinate on the graph, their x-coordinates must sum to 12.

We need to find a pair of points that could be on the graph.

• A) $(5, -9)$ and $(8, -9)$ The y-coordinates are the same ($-9$). Check the sum of the x-coordinates: $5 + 8 = 13$. Since $13 \neq 12$, this pair of points cannot be on the graph.

• B) $(0, 6)$ and $(10, 10)$ The y-coordinates are different. Let's see if both points can satisfy $g(x) = (x-6)^2 + b$ for some value of $b$. For $(0, 6)$: $6 = (0-6)^2 + b \implies 6 = (-6)^2 + b \implies 6 = 36 + b \implies b = -30$. So, $g(x) = (x-6)^2 - 30$. Now check $(10, 10)$ with this function: $10 = (10-6)^2 - 30 \implies 10 = (4)^2 - 30 \implies 10 = 16 - 30 \implies 10 = -14$. This is false, so this pair of points cannot be on the graph.

• C) $(5, 6)$ and $(7, 8)$ The y-coordinates are different. For $(5, 6)$: $6 = (5-6)^2 + b \implies 6 = (-1)^2 + b \implies 6 = 1 + b \implies b = 5$. So, $g(x) = (x-6)^2 + 5$. Now check $(7, 8)$ with this function: $8 = (7-6)^2 + 5 \implies 8 = (1)^2 + 5 \implies 8 = 1 + 5 \implies 8 = 6$. This is false, so this pair of points cannot be on the graph.

• D) $(2, -5)$ and $(10, -5)$ The y-coordinates are the same ($-5$). Check the sum of the x-coordinates: $2 + 10 = 12$. Since $12 = 12$, this pair of points satisfies the symmetry condition. Let's verify by finding $b$. For $(2, -5)$: $-5 = (2-6)^2 + b \implies -5 = (-4)^2 + b \implies -5 = 16 + b \implies b = -21$. So, $g(x) = (x-6)^2 - 21$. Now check $(10, -5)$ with this function: $-5 = (10-6)^2 - 21 \implies -5 = (4)^2 - 21 \implies -5 = 16 - 21 \implies -5 = -5$. This is true. Both points lie on the graph when $b = -21$.

The final answer is $\boxed{\text{D}}$.