Answer to Question #268275 in Calculus for Nicc

Question #268275

Determine the y-intercept, the zeros, the number of turning points, and the behavior of the graph and sketch the graph of the following polynomial functions.



1. P(x)=-2x³-22x²-60x


2. P(x)=6x³+5x²-2x-1


4. P(x)=x⁴+x³-10x²+8




1
Expert's answer
2021-11-21T11:59:55-0500

1. P(x)=2x322x260xP(x)=-2x^3-22x^2-60x

yy -intercept: x=0x=0

P(0)=2(0)322(0)260(0)=0P(0)=-2(0)^3-22(0)^2-60(0)=0

Point (0,0)(0,0)

The graph passes through the origin.

zeros: y=0=>2x322x260x=0y=0=>-2x^3-22x^2-60x=0


2x(x2+11x+30)=0-2x(x^2+11x+30)=0

2x(x+6)(x+5)=0-2x(x+6)(x+5)=0

zeros;6,5,0zeros; -6, -5, 0

Point (6,0)(-6,0), Point (5,0)(-5,0), Point (0,0)(0,0)


y=6x244x60y'=-6x^2-44x-60

y=0=>6x244x60=0y'=0=>-6x^2-44x-60=0

3x2+22x+30=03x^2+22x+30=0

D=(22)24(3)(30)=124D=(22)^2-4(3)(30)=124

x=22±1242(3)=11±313x=\dfrac{-22\pm\sqrt{124}}{2(3)}=\dfrac{-11\pm\sqrt{31}}{3}

There are two turning points.

yy\to \infin as xx\to-\infin

yy\to -\infin as xx\to\infin



2. P(x)=6x3+5x22x1P(x)=6x^3+5x^2-2x-1

yy -intercept: x=0x=0

P(0)=6(0)3+5(0)22(0)1=1P(0)=6(0)^3+5(0)^2-2(0)-1=-1

Point (0,1)(0,-1)


zeros: y=0=>6x3+5x22x1=0y=0=>6x^3+5x^2-2x-1=0


6x3+6x2x2xx1=06x^3+6x^2-x^2-x-x-1=0

6x2(x+1)x(x+1)(x+1)=06x^2(x+1)-x(x+1)-(x+1)=0




(x+1)(6x2x1)=0(x+1)(6x^2-x-1)=0

(x+1)(2x1)(3x+1)=0(x+1)(2x-1)(3x+1)=0


zeros;1,1/3,1/2zeros; -1, -1/3, 1/2

Point (1,0)(-1,0), Point (1/3,0)(-1/3,0), Point (1/2,0)(1/2,0)


y=18x2+10x2y'=18x^2+10x-2

y=0=>18x2+10x2=0y'=0=>18x^2+10x-2=0

9x2+5x1=09x^2+5x-1=0

D=(5)24(9)(1)=61D=(5)^2-4(9)(-1)=61

x=5±612(9)x=\dfrac{-5\pm\sqrt{61}}{2(9)}

There are two turning points.

yy\to -\infin as xx\to-\infin

yy\to \infin as xx\to\infin




4. P(x)=x4+x310x2+8P(x)=x^4+x^3-10x^2+8

yy ​-intercept:x=0x=0

P(0)=(0)4+(0)310(0)2+8=8P(0)=(0)^4+(0)^3-10(0)^2+8=8

Point (0,8)(0,8)


zeros: y=0=>x4+x310x2+8=0y=0=>x^4+x^3-10x^2+8=0


x4x3+2x32x28x2+8=0x^4-x^3+2x^3-2x^2-8x^2+8=0

x3(x1)+2x2(x1)8(x+1)(x1)=0x^3(x-1)+2x^2(x-1)-8(x+1)(x-1)=0




(x1)(x3+2x28x8)=0(x-1)(x^3+2x^2-8x-8)=0

x1=1x_1=1

P(4)=(4)4+(4)310(4)2+8=40>0P(-4)=(-4)^4+(-4)^3-10(-4)^2+8=40>0


P(3)=(3)4+(3)310(3)2+8=28<0P(-3)=(-3)^4+(-3)^3-10(-3)^2+8=-28<0

There is c1(4,3)c_1\in(-4, -3) such that y(c1)=0.y(c_1)=0.


P(1)=(1)4+(1)310(1)2+8=2<0P(-1)=(-1)^4+(-1)^3-10(-1)^2+8=-2<0


P(0)=(0)4+(0)310(0)2+8=8>0P(0)=(0)^4+(0)^3-10(0)^2+8=8>0

There is c2(1,0)c_2\in(-1, 0) such that y(c2)=0.y(c_2)=0.



P(2)=(2)4+(2)310(2)2+8=8<0P(2)=(2)^4+(2)^3-10(2)^2+8=-8<0


P(3)=(3)4+(3)310(3)2+8=26>0P(3)=(3)^4+(3)^3-10(3)^2+8=26>0

There is c3(2,3)c_3\in(2, 3) such that y(c3)=0.y(c_3)=0.

zeros;c1,c2,1,c3.zeros; c_1, c_2, 1, c_3.

Point (1,0)(-1,0), Point (1/3,0)(-1/3,0), Point (1/2,0)(1/2,0)


y=4x3+3x220xy'=4x^3+3x^2-20x

y=0=>4x3+3x220x=0y'=0=>4x^3+3x^2-20x=0

x(4x2+3x20)=0x(4x^2+3x-20)=0

x1=0x_1=0

Or


4x2+3x20=04x^2+3x-20=0

D=(3)24(4)(20)=329D=(3)^2-4(4)(-20)=329

x=3±3292(4)x=\dfrac{-3\pm\sqrt{329}}{2(4)}

There are three turning points.

yy\to\infin ​as xx\to-\infin

yy\to \infin ​as xx\to\infin




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