1. Let f(x) be defined by the linear function graphed below and
g(x) = x2 - 9x + 14.
I
A. Describe the right end behavior of g(x).

1 Let fx be defined by the linear function graphed below and gx x2 9x 14 I A Describe the right end behavior of gx class=

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Answer:

Step-by-step explanation:

A). g(x) = x² - 9x + 14

     Since coefficient of highest degree term (x²) is positive, parabola will open upwards.

    For any parabola opening upwards,

    Right end behavior,

    y → ∞ as x → ∞

B). Let the equation of the linear function is,

    f(x) = mx + b

    Where m = slope of the function

    b = y-intercept

    From the graph attached,

    Slope 'm' = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{-(2+1)}{(0+1)}[/tex]

    m = -3

    b = -1

    Therefore, function 'f' will be,

    f(x) = (-3)x - 1

    f(x) = -3x - 1    

    g(x) = x² - 9x + 14

           = x² - 7x - 2x + 14

           = x(x - 7) - 2(x - 7)

           = (x - 2)(x - 7)

    If h(x) = f(x)g(x)

    h(x) = -(3x + 1)(x -2)(x - 7)

    For h(x) ≥ 0

    -(3x + 1)(x - 2)(x - 7) ≥ 0

     [tex]x\leq -\frac{1}{3}[/tex] Or 2 ≤ x ≤ 7

    Therefore, for 2 ≤ x ≤ 7, h(x) ≥ 0

C). If k(x) = [tex]\frac{h(x)}{(x-2)}[/tex]

    k(x) = [tex]\frac{-(3x+1)(x-2)(x-7)}{(x-2)}[/tex]

    k(x) = -(3x + 1)(x - 7)

    For k(x) = -56

    -(3x + 1)(x - 7) = -56

     3x² -20x - 7 = 56

     3x² - 20x - 63 = 0

     3x² - 27x + 7x - 63 = 0

     3x(x - 9) + 7(x - 9) = 0

     (3x + 7)(x - 9) = 0

     [tex]x=-\frac{7}{3},9[/tex]

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