Question

g(x)=3(1-t)^2-4(1-t)+14
g(1-t)=?

Answer 1

is g(x) like below:
g(x) = 3x^2-4x+14

Answer 2

if yes, the replacing x with 1-t, we get:
\[\large g\left( {1 - t} \right) = 3{\left( {1 - t} \right)^2} - 4\left( {1 - t} \right) + 14\]

Answer 3

then*

Answer 4

please simplify the above expression

Answer 5

I'm having trouble foiling, I think and then simplifying.
I got to
\[3-6t+3t ^{2}-4+4t+14\]

Answer 6

I'm confused about the question
g(x)=3(1-t)^2-4(1-t)+14 means g(x) is constant unless t is an expression of x

Answer 7

if g(x)=a where a is constant then g(1-t)=a also