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Find the Zeros for the polynomial functional and give the multiplicity for each zero. State whether the graph corsses the x-axis or touches the x-axis at each zero!

2006-07-10 16:11:25 · 7 answers · asked by lsubetty 2 in Science & Mathematics Mathematics

7 answers

[1] ROOTS OF THE FUNCTION:


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

f(x) = (x+1/4)^4 (x^2+1)^3
0= (x^2+1)^3
x^2 = -1
x = √(-1)
x = i

[2] GRAPH:

Does the graph touch the axis?
f(0) = (0+1/4)^4 (0^2+1)^3
Graph Touches the graph at: f(0)
f(0) = (1/4)^1/4 +1 ~ 1.707

2006-07-10 17:18:51 · answer #1 · answered by Ernest Maxwell 2 · 2 2

Set each factor = to 0.

x+1/4 =0 > x= - 1/4 Multiplicity is 4 (power on the factor)

X^2 + 1 = 0 > x^2 = -1 > x = +/- i (imaginary solution, i and -i each have multiplicity 3)

The graph just touches the x-axis at 1/4, since that factor has an even power.

2006-07-10 16:20:33 · answer #2 · answered by jenh42002 7 · 0 0

(x + 1/4) = 0, x = -1/4, but since it was power 4, it has multiplicity 4. The graph just touches the x-axis because for even numbered multiplicity, the graph only touches the "zero"

(x^2 + 1) = 0 x = sqrt(-1), with power 3, it has multiplicity 3, but it is imaginary, so it doesn't cross the real axes.
(Normally, odd numbered multiplicities involve the graph crossing the x-axis, thus changing sign)

2006-07-10 16:27:01 · answer #3 · answered by Anonymous · 0 0

x + 1/4 = 0
x = -1/4
This has a multiplicity of 4 (because that is the exponent on (x+1/4))
It touches the x-axis because the multiplicity is even.
(Note: It crosses the x-axis when the multiplicity is odd.)
x^2 + 1 = 0
x^2 = -1
This has imaginary solutions.
(I assume you are only looking for real solutions.)

2006-07-10 16:14:54 · answer #4 · answered by MsMath 7 · 0 0

it really is a quadratic equation: x^2 - 7x = -13 [Make the equation equivalent 0] x^2 - 7x + 13 = 0 [Now, use the quadratic formulation] ax^2 + bx + c =0 x = (-b +/- (b^2 - 4ac)^one million/2)/(2a) x= (Imaginary Numbers) There are not any authentic techniques to the equation; unsolvable to the quantity of authentic techniques. The quadratic formulation is genuine on mine; the variable ^one million/2 is an similar as sq. root. I in basic terms did not favor to discover the sq. root image.

2016-12-10 07:43:30 · answer #5 · answered by ? 4 · 0 0

I know, but I'm not going to tell you. (Yeah, right!) Good luck!

2006-07-10 16:16:37 · answer #6 · answered by NannyMcPhee 5 · 0 0

do ur own homework.

2006-07-10 16:14:17 · answer #7 · answered by Rabi 3 · 0 0

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