Answer to Question #268275 in Calculus for Nicc

Question

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

Answer 1

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

$y$ -intercept: $x=0$

$P(0)=-2(0)^3-22(0)^2-60(0)=0$

Point $(0,0)$

The graph passes through the origin.

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

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

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

$zeros; -6, -5, 0$

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

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

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

$$3x^2+22x+30=0$$

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

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

There are two turning points.

$y\to \infin$ as $x\to-\infin$

$y\to -\infin$ as $x\to\infin$

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

$y$ -intercept: $x=0$

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

Point $(0,-1)$

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

$$6x^3+6x^2-x^2-x-x-1=0$$

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

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

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

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

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

$$y'=18x^2+10x-2$$

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

$$9x^2+5x-1=0$$

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

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

There are two turning points.

$y\to -\infin$ as $x\to-\infin$

$y\to \infin$ as $x\to\infin$

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

$y$ ​-intercept:$x=0$

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

Point $(0,8)$

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

$$x^4-x^3+2x^3-2x^2-8x^2+8=0$$

$$x^3(x-1)+2x^2(x-1)-8(x+1)(x-1)=0$$

$$(x-1)(x^3+2x^2-8x-8)=0$$

$x_1=1$

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

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

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

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

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

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

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

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

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

$zeros; c_1, c_2, 1, c_3.$

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

$$y'=4x^3+3x^2-20x$$

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

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

$$x_1=0$$

Or

$$4x^2+3x-20=0$$

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

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

There are three turning points.

$y\to\infin$ ​as $x\to-\infin$

$y\to \infin$ ​as $x\to\infin$