# Thread: Quick range question: find range of f(x)=(x-5)/X^2-5x+6

1. ## Quick range question: find range of f(x)=(x-5)/X^2-5x+6

Hi all

quick question

given f(x)=(x-5)/X^2-5x+6 find range

EDIT

ok so i think i got it, let me know if this seems right

so write the equation as y=(x-5)/(x^2-5x+6)

simplify y(x^2-5x+6)=(x-5)

range of y will be values where discriminant > or equal to 0

yx^2 - 5xy + 6y = x - 5
yx^2 - 5xy +6y - x + 5 = 0

quadratic is relative to x, where a = y, b = -(5y+1) and c = 6y+5

discriminant = b^2 - 4ac so

(-5y-1)^2 - 4y(6y+5)
25y^2 +5y +5y +1 - 24y^2 - 20y
y^2 -10y + 1 < or equal to 0
y^2 - 10y + 1 = 0

so range = (-infinty, 5-sqrt(24)] u [5+sqrt(24), infinity)

basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated

EDIT

http://www.analyzemath.com/DomainRan..._rational.html

EDIT

Looking for more than just this is how you find range of rational functions lol

2. Originally Posted by frank789
Hi all

quick question

given f(x)=(x-5)/X^2-5x+6 find range

EDIT

ok so i think i got it, let me know if this seems right

so write the equation as y=(x-5)/(x^2-5x+6)

simplify y(x^2-5x+6)=(x-5)

range of y will be values where discriminant > or equal to 0

yx^2 - 5xy + 6y = x - 5
yx^2 - 5xy +6y - x + 5 = 0

quadratic is relative to x, where a = y, b = -(5y+1) and c = 6y+5

discriminant = b^2 - 4ac so

(-5y-1)^2 - 4y(6y+5)
25y^2 +5y +5y +1 - 24y^2 - 20y
y^2 -10y + 1 < or equal to 0
y^2 - 10y + 1 = 0

so range = (-infinty, 5-sqrt(24)] u [5+sqrt(24), infinity)

basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated

EDIT

http://www.analyzemath.com/DomainRan..._rational.html

EDIT

Looking for more than just this is how you find range of rational functions lol
y=(x-5)/(x^2-5x+6)

y=(x-5)/(x^2 - 2x - 3x + 6)

y=(x-5)/[(x - 2)*(x - 3)]

Thus there are two vertical asymptotes at x = 2 & x = 3

continue....

3. Originally Posted by frank789
f(x) = (x - 5)/(x^2 - 5x + 6)

Grouping symbols around denominator are important.

find range

range of y will be values where discriminant > or equal to 0

yx^2 - 5xy + 6y = x - 5

yx^2 - 5xy +6y - x + 5 = 0

discriminant = b^2 - 4ac

(-5y - 1)^2 - 4y(6y + 5)

y^2 - 10y + 1 < or equal to 0

That less-than symbol ought to be a greater-than symbol.

so range = (-infinity, 5 - sqrt(24)] ∪ [5 + sqrt(24), infinity)

However, sqrt(24) can be simplified.

… basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated
Can you be more specific? Are you asking why the Discriminant must be non-negative?

PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.

4. yeah I understand there are vertical asymptotes at 2 and 3 because they are restrictions on the denominator

continuing on then, there must be a vertical asymptote at the restrictions on the inverse of the function.

the discriminant of the equation will tell you the nature of the roots

in this case, the sqrt(b^2 - 4ac) = 0 gives me 2 real solutions

so i can find the solutions to be 5 + or - sqrt(24)

i guess saying that the discriminant > 0 is my restriction so anything between these solutions would not be allowed since it would make the inequality false because it would return the sqrt of a negative

i guess what im getting at is how do I tell the nature of my horizontal asymptotes from the discriminant of the inverse? I really want to avoid "you give me problem i give you an answer" and rather "you gave me this problem this is why it has this solution"

5. Originally Posted by mmm4444bot

However, sqrt(24) can be simplified.

Can you be more specific? Are you asking why the Discriminant must be non-negative?

PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.
unfourtunately, i have to do without calculus XD

im more or less just asking, why can you pull the range from the discriminant? is it because that is the restriction on the function, i.e. values that are not allowed?

6. Originally Posted by mmm4444bot

However, sqrt(24) can be simplified.

Can you be more specific? Are you asking why the Discriminant must be non-negative?

PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.
omg lol the discriminant must be non negative because it is the restriction on the function...thanks for talking me through it :P

7. That's right. We want Real values for the range, so that Discriminant must be non-negative.