# Curve Sketching

1. Aug 8, 2008

1. The problem statement, all variables and given/known data

A curve has equation (x^2 - 2x + 2)/(x^2 + 3x + 9). I need to prove that the range of values of y for this curv eis 2/27 =< y =< 2

I know how it should be worked but I can't understand why it's done that way.

My teacher simply arranged the equation in the form (x^2 + 3x + 9)y = x^2 - 2x + 2 which leads to the equation (y-1)x^2 + (3y+2)x + 9y - 2 = 0 and then said "For range, b^2 - 4ac >= 0"

This is what I'm having trouble understanding. What does the range have to do with b^2 - 4ac >= 0. I want to know the reasoning behind this

Thanks

2. Aug 8, 2008

### HallsofIvy

Staff Emeritus
You teacher used algebraic manipulation to write the equation as $(y-1)x^2 + (3y+2)x + 9y - 2 = 0$, which you can think of as a quadratic equation for x, with the coefficients involving y.

The "range" of a function is the set of all possible y values. Here that means values of y that correspond to real number values of x.

Since that is written as a quadratic equation, use the quadratic formula to solve it.

Remember that the solutions to the quadratic equation $ax^2+ bx+ c= 0$ are given by $(-b\pm \sqrt{b^2- 4ac})/2a$. In particular, those will be real numbers as long as the value inside the squareroot, $b^2- 4ac$ is not negative. Here a= y-1, b= 3y+ 2, and c= 9y- 2.
$$b^2- 4ac= (3y+2)^2- 4(y-1)(9y-2)= 9y^2+ 12y+ 4- 4(9y^2- 11y+ 2)$$
$$= 9y^2+ 12y+ 4- 36y^2+ 44y- 8= -27y^2+ 56y- 4$$

3. Aug 11, 2008