Question

Trace the curve x³ -6x²+11x-y=6

Answer 1

So we can rewrite this curve in another form:

y = x³- 6x² +11x -6

Firstly check if the function is symmetric.

y(-x) = -x³- 6x² -11x -6

We can see that y(x) $\neq$ y(-x), -y(x) $\neq$ y(-x) . This means the curve is neither symmetric with respect to y-axis nor symmetric with respect to the origin.

Let`s find the first derivative:

y'(x) = 3x² - 12x +11

And find extreme points :

3x² - 12x +11 = 0

$D = \sqrt{\smash[b]{(12/2)^2 - 3*11}} = \sqrt{\smash[b]{3}}$

x_1,2= $(6\pm \sqrt{\smash[b]{3}})/3 = 2 \pm 1/\sqrt{\smash[b]{3}}$

We can see two extreme points ($2 -1/\sqrt{\smash[b]{3}}$ , 0.385) and ($2 + 1/\sqrt{\smash[b]{3}}$ , -0.385)

Also our function rises on intervals (-$\infty$, $2 -1/\sqrt{\smash[b]{3}}$ ] , +$\infty$ ) and falls on [$2 -1/\sqrt{\smash[b]{3}},2 +1/\sqrt{\smash[b]{3}}$ ]

Let`s find the second deivative of the function:

y'' (x) = 6x-12

And we get the inflection point x = 2 -> (2,0)

Also this point crosses the X-axis and we can find other cross-points: x³- 6x² +11x -6 = (x-2)*(x² - 4x +3) = (x-1)*(x-2)*(x-3) = 0

And it says that we have 3 points that crosses the X-axis: (1,0), (2,0), (3,0)

Then find point that crosses Y-axis: x=0 -> y(0) = -6

And we have a point that crosses Y-axis: (0, -6).

At last we can draw a sketch of graphic of this function: