Question 24562
If you are given: {{{g(x) = 3x^2-2x+1}}} and are asked to find g(0), you simply replace every x in the original function with 0, thus:
{{{g(0) = 3(0)^2 - 2(0) + 1}}} Simplify.
{{{g(0) = 1}}} ...it's just that easy.

{{{g(3) = 3(3)^2-2(3)+1}}}
{{{g(3) = 3(9)-6+1}}}
{{{g(3) = 27-6+1}}}
{{{g(3) = 22}}}  One more

{{{g(1-t) = 3(1-t)^2 - 2(1-t) + 1}}}
{{{g(1-t) = 3(1-2t+t^2) - 2+2t + 1}}} Simplify.
{{{g(1-t) = 3t^2-6t+3-2+2t+1}}}
{{{g(1-t) = 3t^2-4t+2}}}