Polynomial Equation: Write a quintic with a bounce point at x = 3, zero at x = -1 and x = 4i and -4i which passes through (-2, -125)
Write a quartic with a bounce point at x = 2, a pass through point at x = -5, and passes through (3, -3)
Could I please have some help with the two above problems? I'm having some troubles of where to even start these problems. Any help is much appreciated. Thanks for your time.
Write a quartic with a bounce point at x = 2, a pass through point at x = -5, and passes through (3, -3)
Could I please have some help with the two above problems? I'm having some troubles of where to even start these problems. Any help is much appreciated. Thanks for your time.
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I am assuming that a bounce point is when the graph is tangent to the x-axis at that point; in other words, f(x) has a zero at the bounce point, but it does not cross the x-axis.
(1) In order for a bounce point of f(x) to occur at a certain point, that point must be a zero of f(x) with even multiplicity. The fact that x = 4i, -4i, and -1 are roots takes care of 3 of the 5 roots. In order to get the remaining two roots and to have a bounce point at x = 3, we need x = 3 to be a root of f(x) of multiplicity 3.
Thus, the polynomial takes the form:
f(x) = a(x - 4i)(x + 4i)(x + 1)(x - 3)^2
= a(x^2 + 16)(x + 1)(x - 3)^2, since (x + 4i)(x - 4i) = x^2 - 16i^2 = x^2 + 16.
Since f(x) passes through (-2, -125), we see that:
-125 = a(4 + 16)(-2 + 1)(-2 - 3)^2 ==> a = 1/4.
Thus, the polynomial has equation:
f(x) = (1/4)(x^2 + 16)(x + 1)(x - 3)^2.
(2) I don't think the second question provides enough information. You are given that x = 2 is a root of even multiplicity (since it is a bounce point), a "pass-through" point (where f(x) intersects and crossing the x-axis, indicating a root of ODD multiplicity), and an ordinary point. Since a quartic has four roots, x = 2 is either a root of multiplicity of 2 or 4. However, x = 2 cannot be a root of multiplicity 4 since, if it was, it would take the form:
f(x) = a(x - 2)^4,
which obviously doesn't have x = 5 as a root. Thus, x = 2 must be a root of multiplicity 2 and x = -5 must be a root of multiplicity 1.
Again, a quartic has four roots, but we only have three (2 of multiplicity 2 and -5 of multiplicity 1). We are not given any information about the fourth root. If the fourth root is x = b, then the polynomial takes the form:
f(x) = a(x - b)(x + 5)(x - 2)^2,
but this is obviously not unique.
I hope this helps!
(1) In order for a bounce point of f(x) to occur at a certain point, that point must be a zero of f(x) with even multiplicity. The fact that x = 4i, -4i, and -1 are roots takes care of 3 of the 5 roots. In order to get the remaining two roots and to have a bounce point at x = 3, we need x = 3 to be a root of f(x) of multiplicity 3.
Thus, the polynomial takes the form:
f(x) = a(x - 4i)(x + 4i)(x + 1)(x - 3)^2
= a(x^2 + 16)(x + 1)(x - 3)^2, since (x + 4i)(x - 4i) = x^2 - 16i^2 = x^2 + 16.
Since f(x) passes through (-2, -125), we see that:
-125 = a(4 + 16)(-2 + 1)(-2 - 3)^2 ==> a = 1/4.
Thus, the polynomial has equation:
f(x) = (1/4)(x^2 + 16)(x + 1)(x - 3)^2.
(2) I don't think the second question provides enough information. You are given that x = 2 is a root of even multiplicity (since it is a bounce point), a "pass-through" point (where f(x) intersects and crossing the x-axis, indicating a root of ODD multiplicity), and an ordinary point. Since a quartic has four roots, x = 2 is either a root of multiplicity of 2 or 4. However, x = 2 cannot be a root of multiplicity 4 since, if it was, it would take the form:
f(x) = a(x - 2)^4,
which obviously doesn't have x = 5 as a root. Thus, x = 2 must be a root of multiplicity 2 and x = -5 must be a root of multiplicity 1.
Again, a quartic has four roots, but we only have three (2 of multiplicity 2 and -5 of multiplicity 1). We are not given any information about the fourth root. If the fourth root is x = b, then the polynomial takes the form:
f(x) = a(x - b)(x + 5)(x - 2)^2,
but this is obviously not unique.
I hope this helps!