To find the point that is symmetrical to \((3, -9)\) for the graph of \(y = x^2 - 12x + 18\), we first need to determine the axis of symmetry of the given quadratic function.
The general form of a quadratic equation is given by \(y = ax^2 + bx + c\). Here, \(a = 1\), \(b = -12\), and \(c = 18\). The x-coordinate of the vertex (which is on the axis of symmetry) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \(a\) and \(b\):
\[ x = -\frac{-12}{2 \times 1} = \frac{12}{2} = 6 \]
Thus, the axis of symmetry is the vertical line \(x = 6\).
To find the symmetrical point to \((3, -9)\), we first find the distance from the point \((3, -9)\) to the axis of symmetry \(x = 6\):
\[ \text{Distance} = 6 - 3 = 3 \]
To find the symmetrical point, we move 3 units to the right of the axis of symmetry:
\[ 6 + 3 = 9 \]
Since symmetry about a vertical line does not change the y-coordinate, the y-coordinate remains \(-9\).
Thus, the point symmetrical to \((3, -9)\) for the graph is:
\((9, -9)\)
So the correct response is:
(9,−9)