bagnallshawna
contestada

12. Find all x- and y-intercepts, state the multiplicity of each zero and tell if the graph crosses or
bounces, and state the end behavior Then graph each polynomial by hand,
P(x)=x^2(x-3)^2(x+2)


P(x)=x^4-4x^2

12 Find all x and yintercepts state the multiplicity of each zero and tell if the graph crosses or bounces and state the end behavior Then graph each polynomial class=

Respuesta :

Step-by-step explanation:

12a. To

find x intercepts, set each factor equal to zero

[tex] {x}^{2} = 0[/tex]

[tex]x = 0[/tex]

[tex](x - 3) {}^{2} = 0[/tex]

[tex]x = 3[/tex]

[tex]x + 2 = 0[/tex]

[tex]x = - 2[/tex]

So the x intercepts are( 0,-2,3.)

The multiplicity of 0 and 3 is 2,

The multiplicity of -2 is 1.

The y intercepts are

[tex]0 {}^{2} (0 - 3) {}^{2} (0 + 2) = 0[/tex]

The y intercept is 0.

End Behavior: Since this is a positve 5th root function, as x increases, f(x) increases while as x decreases, f(x) decreases.

The First picture shows the graph of the function. Notice ried the zeroes that have a multiplicity of 2 have a parabolic reflection after thre function reaches their zero.

The zero that have a multiplicity of 1 have just passes through the graph.

12b. We have

[tex] {x}^{4} - 4 {x}^{2} [/tex]

We can easily find y intercept, which is 0.

To find x intercept,

Factor

[tex] {x}^{2} ( {x}^{2} - 4)[/tex]

[tex] {x}^{2} (x + 2)(x - 2) = 0[/tex]

[tex]x = 0[/tex]

[tex]x = 2[/tex]

[tex]x = - 2[/tex]

So the x intercepts are (0,2,-2).

The multiplicity of 0 is 2 while 2 and -2 is 1.

End Behavior: Since this is a positve quartic function, as x increases or decreases , f(x) increases

Look at the second graph,

Ver imagen algebraic12
Ver imagen algebraic12
semsee45

Answer:

[tex]P(x)=x^2(x-3)^2(x+2)[/tex]

y-intercept is when x = 0:

[tex]\implies P(0)=0^2(0-3)^2(0+2)=0[/tex]

Therefore, y-intercept is at (0, 0)

x-intercept(s) is when P(x) = 0

[tex]\implies x^2(x-3)^2(x+2)=0[/tex]

[tex]\implies x^2=0\implies x=0[/tex]

[tex]\implies (x-3)^2=0\implies x=3[/tex]

[tex]\implies (x+2)=0\implies x=-2[/tex]

Therefore, x-intercepts are:

  • (0, 0) with multiplicity 2 → bounces
  • (3, 0)  with multiplicity 2 → bounces
  • (-2, 0) with multiplicity 1 → crosses x-axis

End-behavior:

As the lead coefficient is positive and the lead degree is odd,

[tex]P(x) \rightarrow - \infty \textsf{ as }x\rightarrow - \infty[/tex]

[tex]P(x) \rightarrow + \infty \textsf{ as }x\rightarrow +\infty[/tex]

-------------------------------------------------------------------------------

[tex]P(x)=x^4-4x^2[/tex]

y-intercept is when x = 0:

[tex]\implies P(0)=(0)^4-4(0)^2=0[/tex]

Therefore, y-intercept is at (0, 0)

x-intercept(s) is when P(x) = 0

[tex]\implies x^4-4x^2=0[/tex]

[tex]\implies x^2(x^2-4)=0[/tex]

[tex]\implies x^2(x-2)(x+2)=0[/tex]

[tex]\implies x^2=0 \implies x=0[/tex]

[tex]\implies (x-2)=0 \implies x=2[/tex]

[tex]\implies (x+2)=0 \implies x=-2[/tex]

Therefore, x-intercepts are:

  • (0, 0) with multiplicity 1 → crosses x-axis
  • (2, 0)  with multiplicity 1 → crosses x-axis
  • (-2, 0) with multiplicity 1 → crosses x-axis

End-behavior:

As the lead coefficient is positive and the lead degree is even,

[tex]P(x) \rightarrow + \infty \textsf{ as }x\rightarrow - \infty[/tex]

[tex]P(x) \rightarrow + \infty \textsf{ as }x\rightarrow +\infty[/tex]

Ver imagen semsee45
Ver imagen semsee45
ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas