## Precalculus (6th Edition) Blitzer

The axis of symmetry of the function is $x=-2$ and the required point is $\left( -3,-2 \right)$.
The equation $f\left( x \right)=3{{\left( x+2 \right)}^{2}}-5$ can be written as: $f\left( x \right)=3{{\left( x-\left( -2 \right) \right)}^{2}}+\left( -5 \right)$ Compare the above equation with the standard equation of the parabola $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ , and observe that: $a=3$ , $h=-2,$ and $k=-5.$ The vertex of parabola is $\left( h,k \right)=\left( -2,-5 \right)$. The axis of symmetry of a parabola is given as $x=h$. So, here the axis of symmetry is $x=-2.$ As the graph is symmetric to the above line and the y-coordinate should be the same, the difference of x-coordinates from the axis of symmetry will be the same. The difference of the x-coordinate from the point $\left( -1,-2 \right)$ is $-1-\left( -2 \right)=1$ So the required x-coordinate is $-2-1=-3$ As the y-coordinate is the same, the required point is $\left( -3,-2 \right)$. Hence, the axis of symmetry is $x=-2$ and the required point is $\left( -3,-2 \right)$