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

$f\left(x\right)=x^4-x^3-6x^2+4x+8$.

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

$f\left(x\right)=x^4-x^3-6x^2+4x+8$,

The $y-$intercept is $(0,8)$.

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

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

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

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

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

The graph for the function,

$f\left(x\right)=x^4-x^3-6x^2+4x+8$is,

The turning points of the polynomial function are,

$(-\frac{5}{8}-\frac{1}{8}\sqrt{57},-3.12),(-\frac{5}{8}+\frac{1}{8}\sqrt{57},8.64),$and$(2,0)$.

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$(-1.6,-3.11)$.

The polynomial function increases from $(-\frac{5}{8}-\frac{1}{8}\sqrt{57},-\frac{5}{8}+\frac{1}{8}\sqrt{57})\cup (2,\infty )$.

The polynomial function decreases from,

$(-\infty ,-\frac{5}{8}-\frac{1}{8}\sqrt{57})\cup (-\frac{5}{8}+\frac{1}{8}\sqrt{57},2)$.