I need help with these questions for Working with Standard and Vertex Forms of the Quadratic Function, the first image is the instructions for the questions and the rest of the images have been labelled and put in order of the questions:

9514 1404 393

  3 (7e)  f(x) = -3(x +2)^2 +1; D: (-∞, ∞); R: (-∞, 1]

  4 (8)  g(x) = 4(x -3)^2 -5; x = 3; (3, -5); min: -5; D: (-∞, ∞); R: [-5, ∞)

  5 (10)  A(175) = 31,250 m^2

Step-by-step explanation:

Q3. (7e) f(x) = -3x^2 -12x -11 = -3(x^2 +4x) -11 = -3(x^2 +4x +4) -11 +3(4)

  f(x) = -3(x +2)^2 +1

Domain: all real numbers; Range: y ≤ 1

Expanded: -3(x^2 +4x +4) +1 = -3x^2 -12x -11

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Q4. (8a) g(x) = 4(x^2 -6x) +31 = 4(x^2 -6x +9) +31 -4(9)

  g(x) = 4(x -3)^2 -5

8b) x = 3 . . . axis of symmetry

8c) (3, -5) . . . coordinates of vertex

8d) minimum: -5 (parabola opens upward from vertex, which is the minimum)

8e) any quadratic has a domain of "all real numbers"

8f) the range is upward from the minimum: y ≥ -5

8g) see attached

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Q5. (10) A(x) = 700x -2x^2 = -2(x^2 -350x +30625) +2(30625)

  A(x) = -2(x -175)^2 +31,250

The maximum area that can be enclosed is 31,250 square meters.

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Additional comment

The process of "completing the square" can be done using the following steps. (a) Factor out the leading coefficient from the variable terms; (b) determine the linear term coefficient inside parentheses, and add the square of half of it inside parentheses; (c) add the opposite of the same quantity outside parentheses (make sure the leading coefficient is properly factored in; (d) rewrite the parentheses as a square, and collect terms outside parentheses. In the final form a(x -h)^2 +k, the vertex is (h, k).

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation:

answer: hey there,

the answer you are looking for is a..

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