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FUNCTIONS EXERCISES

Find the range of the linear function y = x+3; the range of any linear function is R(-∞,+∞)

y = x2; R[0,+∞); when the leading term has an even power, the range is limited

y = x2-3; R[-3,+∞); the coefficient of the leading term is positive, therefore the parabola opens upwards

y = 4-x2; R(-∞;4]; the coefficient of the leading term is negative, therefore the parabola opens downwards

y = x2-4x+5; calculating the coordinates of the vertex of the parabola: xv = -b/2a = -(-4)/2⋅1 = 4/2 = 2, yv = 22-4⋅2+5, yv = 4-8+5 = 1, vertex(2,1); the coefficient of the leading term is positive, therefore the parabola opens upwards; R[1,+∞)

y = x3; R(-∞,+∞)

y = x3+5x2-8; for any cubic polynomial the range is always R(-∞,+∞)

y = x5-x3+6x2+8; when the function is a polynomial and the leading term has an odd exponent, the range is always R(-∞,+∞); polynomial functions are continuous functions

y = |x|; the v shape opens upwards because the absolute value function is positive; R[0,+∞)

y = |x|+3; the graph is shifted up by 3 and the v shape opens upwards because the absolute value function is positive; R[3,+∞)

y = |x-2|-3; the graph is shifted right by 2 and down by 3 and the v shape opens upwards because the absolute value function is positive; R[-3,+∞)

y = 2-|x-3|; the graph is shifted up by 2 and right by 3 and the v shape opens downwards because the absolute value function is negative; R(-∞,2]

y = √x; the graph is in the first quadrant, x ≥ 0 and y ≥ 0; R[0,+∞)

y = -√x; the graph is in the fourth quadrant, x ≥ 0 and y ≤ 0; R(-∞,0]

y = √-x; the graph is in the second quadrant, x ≤ 0 and y ≥ 0; R[0,+∞)

y = -√-x; the graph is in the third quadrant, x ≤ 0 and y ≤ 0; R(-∞,0]

y = -√x-3+4; the graph is shifted up by 4 and right by 3; P1(3,4), P2(4,3), P3(7,2); R(-∞,4]