What is the value of k if equation $x^3-3x^2+2=k$ has three real roots and if one real root? I'd like help understanding this question graphically actually. My book finds the derivative of f(x)=$x^3-3x^2+2$ and gets 3x(x-2), thus finding maxima and minima of f(x) as 2 and -2. Then it just says that for three real roots if k belongs to [-2,2] and one real root if k belongs to $(-\infty,-2)$U$(2,\infty)$.Would someone please help me understand this in a graphical,intuitive manner without using discriminant?
 A: Consider $g(x)=x^{3}-3x^{2}=x^{2}(x-3)$, so that $f(x)=g(x)+(2-k)$. The plot of $g(x)$ is as follows;

So we see that $g(x)$ has $2$ roots, one at $x=0$ and the other at $x=3$. It also has a local minimum at $x=2$, with value $f(2)=-4$ and a local maximum of at $x=0$.
Now, $f(x)$ is just a vertical translation of $g(x)$ by $(2-k)$. So for $(2-k)<0$ we have one root, as $g(x)$ is being translated down so its local maximum is below the $x$ axis. For $(2-k)=0$ we have two roots, as $g(x)=f(x)$. For $0<(2-k)<4$ we have three roots, as $g(x)$ has been translated up so that the $x$ axis is between its local minimum and local maximum. For $(2-k)=4$ we have two roots, as $g(x)$ has been translated up until its local minimum is sitting right on the $x$ axis, and finally for $(2-k)>4$ we have one root, as $g(x)$ has been translated until it's local minimum is above the $x$ axis.
As we have $3$ roots for $0<(2-k)<4$, we have by solving the inequality that there are three real roots for $-2<k<2$, i.e. for $k\in(-2,2)$.
As we have a single real, root for $(2-k)<0$ and for $(2-k)>4$, we have upon solving the inequalities that there is one real root for $k>2$ or $x<-2$, i.e. for $k\in (-2,\infty)\cup(2,\infty)$
A: The idea here is that after you calculate your minimum and maximum values, which are $2$ and $0$, respectively, for $x$ and $-2 - K$ and $2 - K$ for $f(x)$, you should try to draw the thing.

Now, for you to have 3 solutions, you need $2 - K \ge 0$ and $-2 - K \leq 0$
Solving this gives you the answer
