Answer to Question #151975 in Analytic Geometry for Sir Jonnsen Dawang

Question

Trace the curve x^3 − 6x^2 + 11x − y = 6.

Answer 1

Equation of the curve is $y = x^3-6x^2+11x-6$

This equation can be written as,

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

When x=0, y=-6.

So, I can directly find the x where y=0,

It will give me, $x= 1,2,3$

So I will have intervals as $(-\infty, 1], [1,2],[2,3],[3,\infty)$

In interval $(-\infty, 1]$, If I put any value from $-\infty$ to 1, value of the y will increase.

In interval $[3, \infty)$ , on increasing value of x, y will increase.

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

Putting $\frac{dy}{dx} =0,$ We get, $x = 1.423,2.577$

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

At x=1.423, $\frac{d^2 y}{dx^2} = 6(1.423)-12 < 0$, Hence it is point of maxima.

At x=2.577, $\frac{d^2 y}{dx^2} = 6x-12 > 0$ , So it is point of minimum.

Using all these properties, We find that curve will look like,