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) = x³ + 6x² + 11x + 6

Answer 1

y-intercept: $x=0$

$$y(0)=(0)^3+6(0)^2+11(0)+6=6$$

Point (0,6)

x-intercepts: $y=0$

$$x^3+6x^2+11x+6=0$$

$$x^3+x^2+5x^2+11x+6=0$$

$$x^2(x+1)+(x+1)(5x+6)=0$$

$$(x+1)(x^2+5x+6)=0$$

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

$$x_1=-3, x_2=-2, x_1=-1$$

There are 3 zeros.

Point (-3,0), point(-2, 0), and point(-1, 0).

$$y'=3x^2+12x+11$$

$$y'=0:3x^2+12x+11=0$$

$D=(12)^2-4(3)(11)=12$

$x_1=\dfrac{-12-\sqrt{12}}{2(3)}=-2-\dfrac{\sqrt{3}}{3}$

$x_2=\dfrac{-12+\sqrt{12}}{2(3)}=-2+\dfrac{\sqrt{3}}{3}$

There are 2 turning points.

$x<-2-\dfrac{\sqrt{3}}{3}, y'>0, y\ increases$

$-2-\dfrac{\sqrt{3}}{3}<x<-2+\dfrac{\sqrt{3}}{3}, y'<0, y\ decreases$

$x>-2+\dfrac{\sqrt{3}}{3}, y'>0, y\ increases$

The function has a local maximum at $x=-2-\dfrac{\sqrt{3}}{3}.$

The function has a local minimum at $x=-2+\dfrac{\sqrt{3}}{3}.$

Sketch the graph