FUNCTIONS EXERCISES

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

y = x^{2}; R[0,+∞); when the leading term has an even power, the range is limited

y = x^{2}-3; R[-3,+∞); the coefficient of the leading term is positive, therefore the parabola opens upwards

y = 4-x^{2}; R(-∞;4]; the coefficient of the leading term is negative, therefore the parabola opens downwards

y = x^{2}-4x+5; calculating the coordinates of the vertex of the parabola: x_{v} = -b/2a = -(-4)/2⋅1 = 4/2 = 2, y_{v} = 2^{2}-4⋅2+5, y_{v} = 4-8+5 = 1, vertex(2,1); the coefficient of the leading term is positive, therefore the parabola opens upwards; R[1,+∞)

y = x^{3}; R(-∞,+∞)

y = x^{3}+5x^{2}-8; for any cubic polynomial the range is always R(-∞,+∞)

y = x^{5}-x^{3}+6x^{2}+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; P_{1}(3,4), P_{2}(4,3), P_{3}(7,2); R(-∞,4]