Find $T(1)$, $T(x)$ and $T(x^{2})$ and $T(ax^2+bx+c)$ Let $$T:P_{3}\rightarrow P_{3}$$ be a linear transformation such that 
$$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1.$$
Find $$T(1)$$, $$T(x)$$ and $$T(x^{2})$$
and $$T(ax^{2}+bx+c)$$.
I will be completely honest and say that I have no idea where to even begin with this problem.
Thank you for the help!

Here's one way:
given that
$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1, \tag{0}$
start with $T(2x^2)$.  We have, by linearity,
$T(2x^2) = 2T(x^2), \tag{1}$
and since we are given
$T(2x^2) = 2x^2 + 3x, \tag{2}$
we see by (1) that
$2T(x^2) = 2x^2 + 3x, \tag{3}$
or
$T(x^2) = x^2 + \dfrac{3}{2}x; \tag{4}$
that covers $T(x^2)$; next, we use
$T(2x^{2}-1)=3x-1, \tag{5}$
together with
$T(2x^2 - 1) = 2T(x^2) -T(1), \tag{6}$
which again follows from the linearity of $T$, to obtain
$2T(x^2) - T(1) = 3x - 1, \tag{7}$
and if we insert the value of $T(x^2)$ from (4) we see that
$2(x^2 + \dfrac{3}{2}x) - T(1) = 3x -1, \tag{8}$
which may be solved for $T(1)$:
$T(1) = 2x^2 + 1. \tag{9}$
Finally, we employ
$T(\dfrac{1}{2}x+2)=2x^{2}+4x-3 \tag{10}$
and the linearity of $T$ once again to see that
$\dfrac{1}{2}T(x) + 2T(1) = 2x^{2}+4x-3; \tag{11}$
substituting $T(1)$ from (9) yields
$\dfrac{1}{2}T(x) + 2(2x^2 + 1) = 2x^2+4x-3, \tag{12}$
and some algebraic fiddling around with (12) further yields
$T(x) = -4x^2 + 8x - 10. \tag{13}$
We have now found $T(1)$, $T(x)$ and $T(x^2)$; finding $T(ax^2 + bx + c)$ is simply one more application of linearity:
$T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1)$
$= a(x^2 + \dfrac{3}{2}x) + b(-4x^2 + 8x - 10) + c(2x^2 + 1)$
$=(a - 4b + 2c)x^2 + (\dfrac{3}{2}a + 8b)x + (c - 10b).  \tag{14}$
The above shows how it can be done in this specific case, in which it is easy to successively isolate $T(1)$, $T(x)$, $T(x^2)$ due to the form of the givens in (0); in the more general case, of which this is but a specific instance, a general linear system in the "variables" $T(1)$, $T(x)$, $T(x^2)$ could be set up and solved using standard linear algebraic techniques.  It boils down to a system of three variables in three unkowns.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
