Functions and Vertex

We need to find ${ g }^{ -1 }(x)$ so we can find $fg[{ g }^{ -1 }(x)]=f(x)$ .

If ${ g }^{ -1 }(x)=y$ , then $g(y)=x$

$\sqrt { \frac { y+2 }{ 2 } } =x\\ \frac { y+2 }{ 2 } ={ x }^{ 2 }\\ y+2=2{ x }^{ 2 }\\ y=2{ x }^{ 2 }-2\\ { g }^{ -1 }(x)={ 2x }^{ 2 }-2$

Now you can find $f(x)$ .

$fg[{ g }^{ -1 }(x)]={ 2x }^{ 2 }-2-9\\ f(x)={ 2x }^{ 2 }-11$

$Vertex=\left( \frac { -b }{ 2a } ,f\left( \frac { -b }{ 2a } \right) \right) \\ a{ x }^{ 2 }+bx+c={ 2x }^{ 2 }-11\\ a=2,\quad b=0,\quad c=-11\\ \\ \frac { -b }{ 2a } =\frac { 0 }{ 4 } =0\\ \\ f\left( \frac { -b }{ 2a } \right) =f\left( 0 \right) =2{ \left( 0 \right) }^{ 2 }-11=-11\\ \\ \therefore Vertex=\left( 0,-11 \right) \\ x-y=0-(-11)=11$