Question

I NEED MORE HELP PLZZ

Answer 1

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.