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find the domain and range in these questions?
use the vertex and intercepts to sketch the graph of each quadratic function. give the equation of the parabola's axis of symmetry.
a- f(x) =2x^2-7x-4
b-f(x) =5-4x-x^2
c-f(x) =6-4x+x^2
1 Answer
- YodaLv 41 decade agoFavorite Answer
(1) f(x) = 2x^2 - 7x - 4 .........f '(x) = 0 = 4x - 7 ........x = 7/4
..f(7/4) = 2(49/16) - (49/4) - 4 = 49/8 - 49/4 - 4 = 49(-1/8) - 4 = -10.125
Vertex = (7/4, -10.125)
........0 = 2x^2 - 7x - 4
........x = [7 ± SQRT(49 + 32)] / 4 = [7 ± SQRT(81)] / 4 = [7 ± 9] / 4
...........= -1/2 ......or..... 4 ..............x-axis intercepts
...span = [4 - (-1/2)] / 2 = 9/4
Axis of symmetry: x = 7/4
Domain: -∞ ≤ x ≤ ∞ ; Range: -10.125 ≤ f(x) ≤ ∞
(2) f(x) = 5 - 4x - x^2 .........f '(x) = 0 = -2x - 4 ........x = -2
....f(-2) = 5 + 8 - 4 = 9
Vertex = (-2, 9)
........0 = 5 - 4x - x^2
........x = [1 ± SQRT(16 + 20)] / 2 = [1 ± SQRT(36)] / 2 = [1 ± 6] / 2
...........= -5/2 ......or..... 7/2 ..............x-axis intercepts
...span = [7/2 - (-5/2)] / 2 = 3
Axis of symmetry: x = 1/2
Domain: -∞ ≤ x ≤ ∞ ; Range: 9 ≤ f(x) ≤ ∞
(3) f(x) = 6 - 4x + x^2 = x^2 - 4x + 6 .........f '(x) = 0 = 2x - 4 ........x = 2
....f(2) = 4 - 8 + 6 = 2
Vertex = (2, 2)
........0 = x^2 - 4x + 6
........x = [4 ± SQRT(16 - 24)] / 2 = [4 ± SQRT(-8)] / 2 = [1 ± 2i SQRT(2)] / 2
...........= [1 - 2i SQRT(2)] / 2 .....or..... [1 + 2i SQRT(2)] / 2 .......x-axis intercepts
...span = [([1 + 2i SQRT(2)] / 2) - ( [1 - 2i SQRT(2)] / 2)] / 2 = 2i SQRT(2)
Axis of symmetry: x = [(1 - 2i SQRT(2)) / 2] + [(4i SQRT(2)) / 2]
..............................= [(1 + 2i SQRT(2)) / 2]
Domain: -∞ ≤ x ≤ ∞ ; Range: 2 ≤ f(x) ≤ ∞
