9514 1404 393
Explanation:
General information
All of these questions have to do with various translations of the square root function. The first problem deals with the basic function. Other problems translate the graph in various ways.
The basic square root function has its vertex at (0, 0) and is defined for x-values (domain) that are 0 or positive. The resulting y-values (range) are also zero or positive.
The basic function intercepts the x- and y-axes only at the point (0, 0), so both intercepts are 0. Unless the function is reflected vertically or horizontally, its minimum will be at the left end of any interval, and its maximum will be at the right end.
The transformation of a function g(x) to f(x) = g(x -a) +b causes it to be translated 'a' units to the right and 'b' units up. For the functions here, that means the vertex is translated from (0, 0) to (a, b). For the functions here, the vertex values are the limits of the domain and range, so those will be altered accordingly.
Problems
1.
a) The graph is shown in the attachment as f1 (orange).
b) vertex: (0, 0); domain: [0, ∞); range: [0, ∞)
c) x-intercept: 0; y-intercept: 0
d) On the interval [1, 4], the minimum is √1 = 1, and the maximum is √4 = 2.
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2.
a) The graph is shown in the attachment as f2 (green). The basic function is translated left 1 and down 2.
b) vertex: (-1, -2); domain: [-1, ∞); range: [-2, ∞)
c) x-intercept: 3; y-intercept: -1 (read from the graph)
d) On the interval, [0.5, 5], the minimum is √1.5 -2 ≈ -0.78; the maximum is √6 -2 ≈ 0.45. Min and max values are shown in the table in the second attachment.
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3.
a) The graph is shown in the attachment as f3 (purple). The basic function is translated up 3.
b) vertex: (0, 3); domain: [0, ∞); range: [3, ∞)
c) x-intercept: (none); y-intercept: 3 (the vertex)
d) On the interval, [2, 4], the minimum is √2 +3 ≈ 4.41; the maximum is √4 +3 ≈ 5. Min and max values are shown in the table in the third attachment.
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4.
a) The graph is shown in the attachment as f4 (red). The leading multiplier of -2 causes the graph to be reflected over the x-axis and vertically expanded by a factor of 2 before being translated right 2 and up 3.
b) vertex: (2, 3); domain: [2, ∞); range: (-∞, 3]. Note the range is all y-values less than or equal to 3 because of the vertical reflection.
c) x-intercept: 4.25; y-intercept: (none). The x-intercept is found by solving ...
0 = -2√(x -2) +3
(-3/-2)² = x-2
2 +2.25 = x = 4.25
d) On the interval, [4, 7], the minimum is -2√(7 -2)+3 = 3-2√5 ≈ -1.47; the maximum is -2√(4-2)+3 = 3-2√2 ≈ 0.17. Min and max values are shown in the table in the second attachment.
