Answer
The axis of symmetry of the function is $x=-2$ and the required point is $\left( -3,-2 \right)$.
Work Step by Step
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 $\left( -3,-2 \right)$
