Answer
See below
Work Step by Step
We are given:
$T(x^2-1)=x^2+x-3\\
T(2x)=4x\\
T(3x+2)=2(x+3)$
Since T is a linear transformation, we have:
$T(x^2)-T(1)=x^2+x-3\\
2T(x)=4x \\
3T(x)+2T(1)=2(x+3)$
Thus:
$T(1)=-2x+3\\
T(x)=2x\\
T(x^2)=x^2-x$
We obtain: $T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)\\
=a(x^2-x)+b(2x)+c(-2x+3)\\
=ax^2+2bx-2cx+3c\\
=ax^2-(a-2b+2c)x+3c$
