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/DomainRange/find_range_rational.html
EDIT
Looking for more than just this is how you find range of rational functions lol
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/DomainRange/find_range_rational.html
EDIT
Looking for more than just this is how you find range of rational functions lol
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