Question

T(-2x²) = -3x² - 3x, T(0.5x + 4) = -2x² + 3x - 1, T(2x² + 1) = -3x - 4. Find T(1), T(x), T(x²), and T(ax² + bx + c), where a, b, and c are arbitrary real numbers. T(1) = -3x² - 6x - 4 T(x) = 20x² + 54x + 30 T(x²) = \frac{3}{2}x² + \frac{3}{2}x T(ax² + bx + c) = \left(-3c + 8b + \frac{3}{2}a\right)x² + \left(-6c + 42b + \frac{3}{2}a\right)x + (-4c + 30b)

          T(-2x²) = -3x² - 3x, T(0.5x + 4) = -2x² + 3x - 1, T(2x² + 1) = -3x - 4.
Find T(1), T(x), T(x²), and T(ax² + bx + c), where a, b, and c are arbitrary real numbers.
T(1) = -3x² - 6x - 4
T(x) = 20x² + 54x + 30
T(x²) = \frac{3}{2}x² + \frac{3}{2}x
T(ax² + bx + c) = \left(-3c + 8b + \frac{3}{2}a\right)x² + \left(-6c + 42b + \frac{3}{2}a\right)x + (-4c + 30b)

T(-2x²) = -3x² - 3x, T(0.5x + 4) = -2x² + 3x - 1, T(2x² + 1) = -3x - 4.
Find T(1), T(x), T(x²), and T(ax² + bx + c), where a, b, and c are arbitrary real numbers.
T(1) = -3x² - 6x - 4
T(x) = 20x² + 54x + 30
T(x²) = (3)/(2)x² + (3)/(2)x
T(ax² + bx + c) = (-3c + 8b + (3)/(2)a)x² + (-6c + 42b + (3)/(2)a)x + (-4c + 30b)

Added by Christopher B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Manuel Vasquez

Step 1

Because T is linear it suffices to determine T(1), T(x), T(x^2); then for any a,b,c we have T(a x^2 + b x + c) = a T(x^2) + b T(x) + c T(1). Show more…

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Please do the last one number (4). Note: Please be correct because I will check the answer and most of the time it's going to be incorrect but I am writing this anyway to caution anyone that I will upvote if the answer is correct and downvote for wrong answers. T-2x^2=-3x-3xT0.5x+4=-2x+3x-1, T2x+1=-3x-4. Find T1, T, T2, and Tax+bx+c, where a, b, and c are arbitrary real numbers. T(1) -32-6x-4 Tx= 202+54r+30 T(x^2) 3.2 3 T(ax^2+bx+c) 3c+8b+ Note: Youcar