The domain is all values of y where y is a defined function.
If the denominator is equal to 0, the function is typically undefined. So here, when:
x+3=0, the function is undefined.
Therefore, at x=-3, the function is undefined.
So, the domain is stated as (-oo,-3)uu(-3,oo).
The range is all possible values of y. It is also found when the discriminant of the function is less than 0.
To find the discriminant (Delta), we must make the equation a quadratic equation.
y=(x^2-x-1)/(x+3)
y(x+3)=x^2-x-1
xy+3y=x^2-x-1
x^2-x-xy-1-3y=0
x^2+(-1-y)x+(-1-3y)=0
This is a quadratic equation where a=1, b=-1-y, c=-1-3y
Since Delta=b^2-4ac, we can input:
Delta=(-1-y)^2-4(1)(-1-3y)
Delta=1+2y+y^2+4+12y
Delta=y^2+14y+5
Another quadratic expression, but here, since Delta>=0, it is an inequality of the form:
y^2+14y+5>=0
We solve for y. The two values of y we get will be the upper and lower bounds of the range.
Since we can factor ay^2+by+c as (y-(-b+sqrt(b^2-4ac))/(2a))(y-(-b-sqrt(b^2-4ac))/(2a)), we can say, here:
a=1, b=14, c=5. Inputting:
(-14+-sqrt(14^2-4*1*5))/(2*1)
(-14+-sqrt(196-20))/2
(-14+-sqrt(176))/2
(-14+-4sqrt(11))/2
+-2sqrt(11)-7
So the factors are (y-(2sqrt(11)-7))(y-(-2sqrt(11)-7))>=0
So y>=2sqrt(11)-7 and y<=-2sqrt(11)-7.
In interval notation we can write the range as:
(-oo,-2sqrt(11)-7]uu[2sqrt(11)-7,oo)
