Analyze each polynomial function using Steps $1$through $8$:

$f\left(x\right)=x^4+x^3-3x^2-x+2$.

The polynomial function $f$is of degree$4$. The graph behaves like$y=x^4$for large values of$\left|x\right|$.

Setting $y=0$, for the equation$f\left(x\right)=x^4+x^3-3x^2-x+2$,

The $y-$intercept is $(0,2)$.

The$x-$intercept is$(-2,0),(-1,0),$and$(1,0)$.

The zeros of $f$are $-2,-1,$and $1$.

The zero $-2$is a zero of the multiplicity $1$, so the graph of $f$crosses the $x-$axis at $x=-2$.

The zero $-1$is a zero of the multiplicity $1$, so the graph of$f$crosses the $x-$axis at $x=-1$.

The zero$1$is a zero of the multiplicity$2$, so the graph of$f$crosses the$x-$axis at$x=1$.

The graph for the function $f\left(x\right)=x^4+x^3-3x^2-x+2$is,

The turning points of the polynomial function are,

$(-\frac{7}{8}-\frac{1}{8}\sqrt{33},-1.62),(-\frac{7}{8}+\frac{1}{8}\sqrt{33},2.08),$and$(1,0)$.

The graph of the function by hand is,

The domain of the function is $\mathrm{ℝ}$which is $(-\infty ,\infty )$.

The range of the function is$[-1.62,\infty )$.

The polynomial function increases from $(-\frac{7}{8}-\frac{1}{8}\sqrt{33},-\frac{7}{8}+\frac{1}{8}\sqrt{33})\cup (1,\infty )$.

The polynomial function decreases on$(-\infty ,-\frac{7}{8}-\frac{1}{8}\sqrt{33})\cup (-\frac{7}{8}+\frac{1}{8}\sqrt{33},1)$.