# Thread: Sketching a curve. Not sure where does this transformation comes from

1. ## Sketching a curve. Not sure where does this transformation comes from

Not sure where last two equasions comes from. Question is rewritten from a book preparing to A-levels. Thanks for help in advance.

Sketch the curve given by y=(3x+3)/(x(3-x)) ***(1)***

"Range of values of y":

from (1): 3xy - (x^2)y = 3x +3 ***(2)***

from (2): (x^2)y + 3x(1-y)+3=0 ***(3)***

so for y=0 => 3x+3=0 => x=-1

Now for y!=-1 this is a quadratic in x.

Thus, for real x,

[3(1-y)]^2 - 4y(3)>= 0
i.e. (3y-1)(y-3) >= 0

...question continues with sketching the curve and is easy apart from last two lines.

Kind regards,

2. ## Re: Sketching a curve. Not sure where does this transformation comes from

Not sure where last two equasions comes from.
The last two equations come from the discriminant of the quadratic formula, $b^2-4ac$

from (2): (x^2)y + 3x(1-y)+3=0
$a = y$

$b = 3(1-y)$

$c = 3$

$b^2 - 4ac = [3(1-y)]^2 - 4(y)(3)$

3. ## Re: Sketching a curve. Not sure where does this transformation comes from

A few things.
The restriction on the domain of x is x != 0,3 so when you say "Thus, for real x" this isn't true.

Second, when you say y != -1 this is a quadratic in x, it's also a quadratic in x when y = -1, namely -x^2 + 6x + 3 = 0 which has no real roots.
It is a quadratic for all values y except for y=0, if the definition of quadratic is that the highest degree with a non-zero coefficient is 2.

The last two lines refer to the determinant of quadratic (3). ie. D = b^2 - 4ac, where a is the coefficient of x^2, b is the coefficient of x, and c is the constant. If D >= 0 this means the quadratic has real roots. (If 0, repeated real roots)
a = y, the coefficient of x^2
b = 3(1-y), the coefficient of x
c = 3

edited because tex tags didn't format.