Grade 12 math question?

write the equation of a member of a quartic family of functions with -5,-1,7/2, and 3 as zeroes and (-2, 25) is a point on the graph.

I have this so far: (x+5) (x+1)(2x-7)(x-3)

3 Answers

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  • 9 years ago
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    f(x) = a(x+5)(x+1)(2x-7)(x-3)

    so, f(-2) = a(3)(-1)(-11)(-5) = 25

    => -165a = 25

    i.e. a = -25/165 = -5/33

    so, f(x) = (-5/33)(x+5)(x+1)(2x-7)(x-3)

    :)>

  • Anonymous
    9 years ago

    First plug in the zeros:

    ie. If x = -5 is a zero, then (x + 5) is a factor, so:

    y = (x + 5)(x + 1)(x - 7/2)(x + a)

    To solve for a, use the point which they gave you.

    When x = -2, y = 25:

    25 = (-2 + 5)(-2 + 1)(-2 - 7/2)(-2 + a)

    (3)(-1)(-11/2)(a - 2) = 25

    33/2(a - 2) = 25

    a - 2 = 50 / 33

    a = 2 + 50 / 33

    a = (66 + 50) / 33

    a = 116 / 33

    So the quartic is:

    y = (x + 5)(x + 1)(x - 7/2)(x + 116/33)

  • Good! Now put y= what you have so far. multiply rhs with a constant k. put in values of x=-2 and y=25, and sole to find k. that will be your equation.

    In this case, k=-5/33.

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