PLEASE HELP!!! How does the multiplicity of a zero affect the graph of the polynomial function?

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Step-by-step explanation:

Consider a zero at x=1.

If it has multiplicity 1, the corresponding factor is (x -1). Now consider the sign of that factor for x = 1.1. The value is (1.1 -1) = 0.1, a positive number.

When the value of x is 0.9, the factor is (0.9 -1) = -0.1, a negative number.

This tells you that values to the left of a zero make the factor(s) associated with that zero be negative. For x-values to the right of the zero, the factor(s) will be positive.

If the multiplicity is even, the product of those factors is positive on either side of the zero. (A negative number to an even power is a positive number.) That is, the graph will not change sign from one side of an even-multiplicity factor to the other. At an odd-multiplicity factor, the graph changes sign.

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For the problem at hand, the only even-multiplicity factor is (x -1), so the graph will not change sign at x=1. It will touch the axis (be tangent to the x-axis), but will not cross. The other zeros have odd multiplicity (1 or 3), so the graph will change sign there.

It is worthwhile to note that the higher the multiplicity, the "flatter" the graph is at the zero. You can see this in the attachment by comparing the shape of the graph at x=6 with the shape at x=2 (also an upward zero crossing).

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: