Given that g(x)=3x^2-2x+2 find each of the following. a) g(0) b) g(-2) c) g(2) d) g(-x) e) g(1-t) a) Select the correct choice below and, if necessary, fill in the answer box to co A. g(0)=square (Simplify your answer.) B. The value g(0) does not exist. b) Select the correct choice below and, if i necessary, fill in the answer box to co A. q(-2)=square (Simplify your answer.)

a) To find $g(0)$, we substitute $x=0$ into the function $g(x)$.<br /><br />$g(0) = 3(0)^{2} - 2(0) + 2 = 0 - 0 + 2 = 2$<br /><br />So, the correct choice is A. $g(0)=2$.<br /><br />b) To find $g(-2)$, we substitute $x=-2$ into the function $g(x)$.<br /><br />$g(-2) = 3(-2)^{2} - 2(-2) + 2 = 3(4) + 4 + 2 = 12 + 4 + 2 = 18$<br /><br />So, the correct choice is A. $g(-2)=18$.<br /><br />c) To find $g(2)$, we substitute $x=2$ into the function $g(x)$.<br /><br />$g(2) = 3(2)^{2} - 2(2) + 2 = 3(4) - 4 + 2 = 12 - 4 + 2 = 10$<br /><br />So, $g(2)=10$.<br /><br />d) To find $g(-x)$, we substitute $x=-x$ into the function $g(x)$.<br /><br />$g(-x) = 3(-x)^{2} - 2(-x) + 2 = 3x^{2} + 2x + 2$<br /><br />So, $g(-x) = 3x^{2} + 2x + 2$.<br /><br />e) To find $g(1-t)$, we substitute $x=1-t$ into the function $g(x)$.<br /><br />$g(1-t) = 3(1-t)^{2} - 2(1-t) + 2$<br /><br />Expanding the expression:<br /><br />$g(1-t) = 3(1 - 2t + t^{2}) - 2(1 - t) + 2$<br /><br />$g(1-t) = 3 - 6t + 3t^{2} - 2 + 2t + 2$<br /><br />$g(1-t) = 3t^{2} - 4t + 3$<br /><br />So, $g(1-t) = 3t^{2} - 4t + 3$.