Answer
a.) The graph fall to the left and rises to the right.
b.) $x= -2, -1, 1$. The function crosses the x-axis at the zeros.
c.) $-2$
d.) Neither.
e.) See attached graph.
Work Step by Step
$f(x)=x^3+2x^2-x-2$
a.) The degree of the function is 3, which is an odd number. Odd-degree functions have graphs with opposite behavior at each end.
The leading coefficient is 1, which is a positive number.
b.) Find the zeros.
$(x^3+2x^2)+(-x-2)$
$x^2(x+2)-1(x+2)$
$(x+2)(x^2-1)$
$(x+2)(x+1)(x-1)$
$x+2=0 \quad or \quad x+1 = 0 \quad or \quad x-1=0$
$x=-2, \quad x=-1, \quad x=1$
The multiplicity of the zeros are odd. This means that the function will cross the x-axis at them.
c.) $f(0)$
$f(0)=(0)^3+2(0)^2-0-2=-2$
y-intercept $=-2$
d.) $f(-x)$
$f(-x)=(-x)^3+2(-x)^2-(-x)-2= -x^3+2x^2+x-2$
$f(-x)\ne f(x), \quad and \quad f(-x)\ne -f(x)$
e.) Max. turning points= 2
Degree= 3
