6.
So for this, since the circle is open, you are going to be replacing x in g(x) with -4x + 1. It will be solved as such:
[tex] g(h(x))=(-4x+1)^2+1\\ g(h(x))=16x^2-8x+1+1\\ g(h(x))=16x^2-8x+2 [/tex]
8.
So firstly, we will need to solve g(f(x)) before we can do h(g(f(x))). So replace x in g(x) with x - 3:
[tex] g(f(x))=(x-3)^2+1\\ g(f(x))=x^2-6x+9+1\\ g(f(x))=x^2-6x+10 [/tex]
So now that we know what g(f(x)) is, we can solve for h(g(f(x))). To do this, replace x in h(x) with x^2 - 6x + 10:
[tex] h(g(f(x)))=-4(x^2-6x+10)+1\\ h(g(f(x)))=-4x^2+24x-40+1\\ h(g(f(x)))=-4x^2+24x-39 [/tex]
In short, h(g(f(x))) = -4x^2 + 24x - 39.
