Answer:
1A: all real numbers
1B: x = {-3, 0, +18}
1C: (-3, 0) U [3, 18)
2A: h(x) = -1/160x^2 +1/2x
2B: focus: (40, -30); directrix: y = 50; axis of symmetry: x = 40
Step-by-step explanation:
1A:
The graph is attached. The domain is all real numbers.
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1B:
The x-intercepts for x < 3 are the zeros of the quadratic:
x^3 -9x = 0
x(x^2 -9) = 0
x(x +3)(x -3) = 0 ⇒ x = 0, -3, +3 . . . . x=3 is not in the domain of this piece
The x-intercepts for x ≥ 3 are the zeros of the log function:
-log4(x -2) +2 = 0
log4(x -2) = 2 . . . . . . add log4(x -2)
x -2 = 4^2 . . . . . . . . . take the anti-log
x = 2 +16 = 18 . . . . . . add 2
The x-intercepts of g(x) are x = {-3, 0, +18}.
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1C:
The graph shows the function is positive on the intervals ...
(-3, 0) U [3, 18)
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2A:
The vertex of the curve is given as (40, 10). The vertical scale factor will make the function 0 at x=0.
h(x) = a(x -40)^2 +10
h(0) = 0 = a(-40)^2 +10 = 1600a +10
a = -10/1600 = -1/160 = -1/(4·40)
The equation of the parabola is ...
y = -1/(160)(x -40)^2 +10
In standard form, this is ...
y = -1/160x^2 +1/2x
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2B:
When we computed the leading coefficient (a) in Part A, we found it to be -1/(4·40). This corresponds to 1/(4p) where p=-40 is the focus to vertex distance. The negative sign means the focus is 40 units below the vertex, at (40, -30). The corresponding location of the directrix is 40 units above the vertex, at y=50.
Of course the axis of symmetry is the vertical line through the vertex: x = 40. These are shown in the second attachment.
