Question

let g(x)=x-3 and h(x)=x^2+6 find (h o g) (1)

Answer 1

$h \circ g=(x-3)^2+6=x^2-6x+9+6=x^2-6x+15$

Answer 2

$g(x)=x-3 \ \ \ and\ \ \ h(x)=x^2+6 \\\\(h \circ g) (1)=h\bigg (g(1)\bigg)=h\bigg(1-3\bigg)=h(-2)=(-2)^2+6=4+6=10$