Consider the transformation T(ax^2 + bx + c) = 2ax + b of P2 → P1.
(a) Find T((ax^2 + bx + c) + (px^2 + qx + r)).
(b) Find T(k(ax^2 + bx + c)).
(c) Find the image of 3x^2 − 2x + 1.
(d) Determine another element of P2 that has the same image.
(a) Find T((ax^2 + bx + c) + (px^2 + qx + r)).
(b) Find T(k(ax^2 + bx + c)).
(c) Find the image of 3x^2 − 2x + 1.
(d) Determine another element of P2 that has the same image.
a) 2ax+2px+b+q
b) k(2ax+b)
c) 6x-2
d) 3x^2-2x+6
(a) To find T((ax^2 + bx + c) + (px^2 + qx + r)), we first add the two polynomials together:
(ax^2 + bx + c) + (px^2 + qx + r) = (a + p)x^2 + (b + q)x + (c + r)
Now we can apply the transformation T to this sum:
T((ax^2 + bx + c) + (px^2 + qx + r)) = 2(a + p)x + (b + q)
So, the result is 2(a + p)x + (b + q).
(b) To find T(k(ax^2 + bx + c)), we multiply the polynomial by the constant k:
k(ax^2 + bx + c) = kax^2 + kbx + kc
Now we can apply the transformation T to this polynomial:
T(k(ax^2 + bx + c)) = 2kax + kb
So, the result is 2kax + kb.
(c) To find the image of 3x^2 − 2x + 1, we substitute the polynomial into the transformation T:
T(3x^2 − 2x + 1) = 2(3x^2) + (-2) = 6x^2 - 2
So, the image of 3x^2 − 2x + 1 is 6x^2 - 2.
(d) To determine another element of P2 that has the same image as 3x^2 − 2x + 1, we need to find a polynomial in P2 that, when transformed by T, gives the same result:
T(ax^2 + bx + c) = 6x^2 - 2
Comparing the coefficients of the transformed polynomial with the original polynomial, we have:
2ax = 6x^2
b = 0
c = -2
From the first equation, we can see that a = 3. So, one possible polynomial is:
3x^2 - 2
This polynomial has the same image as 3x^2 − 2x + 1 under the transformation T.
To find the answers to the given questions, we need to understand the transformation T and how it operates on polynomials in P2. Let's go through each question step by step.
(a) Find T((ax^2 + bx + c) + (px^2 + qx + r)):
To find the result of this expression under the transformation T, we apply T to each term separately and then combine the results.
T(ax^2 + bx + c) + T(px^2 + qx + r)
= (2ax + b) + (2px + q)
= (2ax + 2px) + (b + q)
= 2x(a + p) + (b + q)
So, T((ax^2 + bx + c) + (px^2 + qx + r)) = 2x(a + p) + (b + q).
(b) Find T(k(ax^2 + bx + c)):
In this case, we have a scalar k multiplying the polynomial (ax^2 + bx + c). Applying T, we distribute k across each term as follows:
T(k(ax^2 + bx + c))
= T(kax^2 + kbx + kc)
= 2kax + kb + 0
= 2kax + kb
So, T(k(ax^2 + bx + c)) = 2kax + kb.
(c) Find the image of 3x^2 − 2x + 1:
The image of a polynomial under the transformation T is obtained by directly applying T to the polynomial. Let's substitute the given polynomial into the expression for T:
T(3x^2 − 2x + 1)
= 2(3x) + (-2)
= 6x - 2
Therefore, the image of 3x^2 − 2x + 1 under T is 6x - 2.
(d) Determine another element of P2 that has the same image:
To find another element of P2 that has the same image as the polynomial 3x^2 − 2x + 1, we need to solve for a polynomial of the form ax^2 + bx + c such that T(ax^2 + bx + c) equals 6x - 2.
Let's equate the expressions for T(ax^2 + bx + c) and 6x - 2:
2ax + b = 6x
b = 0 (to match the constant term)
2a = 6 (to match the x coefficient)
a = 3
Therefore, another element of P2 that has the same image as the polynomial 3x^2 − 2x + 1 under T is 3x^2.