Question

Trace the curve y

2 = (x − 1)(x − 2)(x − 3).

Answer 1

Equation of the curve : $\pmb {y=(x-1)(x-2)(x-3)}$

$y=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$

Differentiating w.r.t x,

$\frac{dy}{dx}=3x^2-12x+11$ (1)

Symmetry :

The curve is not symmetric w.r.t. any line or axis.

Region :

$y=(x-1)(x-2)(x-3)\\$

$y$ is defined $\forall x\in \R$

$\therefore$ Domain of $y=(-\infty,\infty)$

Now from (1), $\frac{dy}{dx}$ exists $\forall x\in \R$

So, $y=f(x)$ is continuous $\forall x\in\R$ (2)

Now,

$\;\;\;\;\;\;\;\;x\to \infty\Rightarrow y\to \infty\\ and, \;y\to -\infty\Rightarrow y\to-\infty$

So from (2), by intermediate-value theorem,

$\forall y\in \R\;\;\exists x\in\R \ni y=(x-1)(x-2)(x-3)$

$Range \; of \;y=(-\infty,\infty)$

$\therefore x\in(-\infty,\infty),\;\; y\in (-\infty,\infty)$

Point of Intersection :

Putting x=0, we get y=-6

and Putting y=0, we get $x=1,2,3$

$\therefore$ The curve cuts x-axis at $(1,0),(2,0),(3,0)$ and,

the curve cuts y-axis at $(0,-6)$

Tangent at Origin :

The curve does not go through the origin. So there is no tangent at the origin.

Asymptotes :

Coefficient of highest degree of $x$ and $y$ are both constant. So, There does not exist any horizontal or vertical asymptote.

Putting $y=mx+c,\\$

$mx+c=x^3-6x^2+11x-6\\ \Rightarrow x^3-6x^2+(11-m)x-(c+6)=0$

$\Rightarrow$ Coff. of $x^3=1=Constant$

So There does not exist any oblique asymptote either.

Derivatives :-

$\frac{dy}{dx}=3x^2-12x+11$

Differentiating w.r.t x,

$\frac{d^2y}{dx^2}=6x-12=6(x-2)$

Now

$\frac{dy}{dx}=0\\ \Rightarrow 3x^2-12x+11=0\\ \Rightarrow x=\frac{12\pm \sqrt{12^2-4\times3\times11}}{6}=\frac{6\pm\sqrt3}{3}=2\pm\frac{1}{\sqrt3}$

$\frac{d^2y}{dx^2}_{[x=2+1/\sqrt3]}=\frac{6}{\sqrt3}>0$

$\therefore$ There is a local minimum at $x=2+1/\sqrt3$

and,

$\frac{d^2y}{dx^2}_{[x=2-1/\sqrt3]}=-\frac{6}{\sqrt3}<0$

$\therefore$ There is a local minimum at $x=2-1/\sqrt3$

Graph :