91Ó°ÊÓ

$f\left(x\right)=-x^5+5x^4+4x^3-20x^2$

$$\begin{gathered} f\left(x\right)=-x^5+5x^4+4x^3-20x^2 \\ =-x^2(x^3-5x^2-4x+20) \\ =-x^2\left\{x^2\right(x-2)--3x(x-2)-10(x-2\left)\right\} \end{gathered}$$

localid="1650169764787" $$\begin{gathered} =-x^2(x-2)\{x^2-3x-10\} \\ =-x^2(x-2)\{x^2-5x+2x-10\} \\ =-x^2(x-2)\left\{x\right(x-5)+2(x-5\left)\right\} \\ =-x^2(x-2)(x-5)(x+2) \end{gathered}$$

$f\left(x\right)=-x^5+5x^4+4x^3-20x^2$

Clearly, as $x→∞,f\left(x\right)→-∞$

and as$x→-∞,f\left(x\right)→∞$

For $y-$intercept

$f\left(0\right)=0$i.e.$(0,0)$

and for $x-$intercept

$f\left(x\right)=0$

or, $-x^2(x-2)(x-5)(x+2)=0$

Then $x=-2,0,2,5$

Then $x-$intercepts are$(-2,0),(0,0),(2,0),(5,0)$

Now, we have to find points on the function around x & y intercept

For $x=-2$

For$x=0$

For $x=2$

For $x=5$

The turning points of the function in the graph are$(-1.451,-25.732),(0,0),(1.341,-14.487),(4.11,193.89)$

The domain and the range of the given function is the set of all real numbers.

The function is increasing on $-1.451<x≤0\&1.341<x≤4.11$.

The function is decreasing on$-∞<x≤-1.451,0<x≤1.341\&4.11<x<∞$.