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Determine the equation of any asymptotes?

a) f(x)=(3x-7)/(6x+3)

b) g(x)=(2x-5)/(2x^2-8)

c) h(x)=(3x^2-4x+2)/((1/2 x)^2+2x-1)

2 Answers

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  • cidyah
    Lv 7
    4 years ago

    a)

    f(x) = (3x-7) /(6x+3)

    Vertical asymptote :

    6x+3 = 0

    6x=-3

    x=-1/2

    Horizontal asymptote:

    lim x-->infinity (3x-7) /(6x+3) = lim x-->inf x(3-7/x) / (x(6+3/x)) = lim x-->inf (3-7/x)/(6+3/x)

    = 3/6 = 1/2

    f(x) = 1/2 is the horizontal asymptote

    b)

    g(x) = 2(x-5/2) / (2(x^2-4))

    y = (x-5/2) /((x+2)(x-2))

    Vertical asymptotes are x=-2, x=2

    Horizontal asymptote: since the denominator as a larger exponent than that of the numerator, y goes to 0 as x approaches infinity

    g(x) = 0 is the horizontal asymptote

    c)

    h(x) = 4(3x^2-4x+2) / (x^2+2x-1)

    h(x) = (12x^2-16x+8) / (x^2+2x-1)

    h(x) = x^2 (12-16/x +8/x^2) / x^2 (1+2/x-1/x^2)

    h(x) = (12-16/x+8/x^2) /(1+2/x-1/x^2)

    lim x-->inf (12-16/x+8/x^2) /(1+2/x-1/x^2) = 12/ 1 = 12

    h(x) = 12 is the horiontal asymptote

    Vertical asymptote is x^2+2x-1 = 0

    This equation is of form ax^2+bx+c

    a = 1 b = 2 c = -1

    x=[-b+/-sqrt(b^2-4ac)]/2a]

    x=[-2 +/-sqrt(2^2-4(1)(-1)]/(2)(1)

    discriminant is b^2-4ac =8

    x=[-2 +√(8)] / (2)(1)

    x=[-2 -√(8)] / (2)(1)

    x = -1 + √2 and x= -1 - √2 are vertical asymptotes.

  • ?
    Lv 7
    4 years ago

    a) Vertical asymptote at x = -1/2 (set 6x + 3 = 0 to see why). Horizontal asymptote at y = 1/2 (3x / 6x ---> 3/6 = 1/2; take the limit as x approaches negative or positive infinity).

    b) Vertical asymptotes at x = -2 and x = 2 (2x^2 - 8 = 0; 2x^2 = 8; x^2 = 4...you should be able to obtain 2 and -2 from there). Horizontal asymptote at y = 0 (if the degree of the numerator is less than that of the denominator, the limit is 0 and the horizontal asymptote is y = 0).

    c) As written, ((1/2)x)^2 = (1/4)x^2, so:

    (1/4)x^2 + 2x - 1 = 0

    x^2 + 8x - 4 = 0

    Discriminant: 8^2 - 4(1)(-4) = 64 + 16 = 80. Two irrational vertical asymptotes. Using the whole quadratic formula yields x = (-8 +/- 4 sqrt(5)) / 2 ---> x = -4 - 2 sqrt(5) and x = -4 + 2 sqrt(5). Those are the vertical asymptotes.

    Horizontal asymptote is y = 12 (3x^2 / (1/4)x^2 ---> 3 / (1/4) = 3 * 4 = 12).

    ***********************************************************************************************

    POSSIBLE EDIT FOR c): If the denominator is supposed to be (1/2)x^2 + 2x - 1, then:

    (1/2)x^2 + 2x - 1 = 0

    x^2 + 4x - 2 = 0

    Quadratic formula yields (-4 +/- sqrt(4^2 - 4(1)(-2)) / 2(1) = (-4 +/- sqrt(24)) / 2 ---> x = -2 + sqrt(6) and x = -2 - sqrt(6). Again, two irrational vertical asymptotes. Horizontal asymptote is y = 6 (3 / (1/2) = 3 * 2 = 6).

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