## Precalculus (6th Edition) Blitzer

The axis of symmetry of the function is $x=3$ and the required point is $\left( 0,11 \right)$
The equation $f\left( x \right)={{\left( x-3 \right)}^{2}}+2$ can be written as: $f\left( x \right)={{\left( x-3 \right)}^{2}}+2$ Compare the above equation with the standard equation of a parabola $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ , and observe that: $a=1$ , $h=3,$ and $k=2.$ The vertex of a parabola is $\left( h,k \right)=\left( 3,2 \right)$ The axis of symmetry of a vertical line is given as $x=h$. So, here the axis of symmetry is $x=3.$ The graph is symmetric to the above line and the y-coordinate should be the same, so the difference of the x-coordinates from the axis of symmetry will be the same. Now, the difference of the x-coordinate from the point $\left( 6,11 \right)$ is $6-\left( 3 \right)=3$ Thus, the required x-coordinate is $3-3=0$ As the y-coordinate is the same, the required point is $\left( 0,11 \right)$. Hence, the axis of symmetry is $x=3$ and the required point is $\left( 0,11 \right)$.