Ira S. answered • 08/09/14
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I'm not sure what this problem is really asking. I wish I could just talk to you for clarification.
Are you studying rational functions? Do you know about vertical and horizontal asymptotes? Is this
f(x)=(x^2 +2x+a)/(x^2 +4x +3a)
Ira S.
I think what I'm thinking is way too complicated. so I'll start you off with this. If this fraction can be every real number, then it can be the specific number 0. A fraction is 0 when the numerator is 0. So x^2 +2x +a = 0 when x= (-2+-sqrt 4- 4a)/2 using quadratic formula. In order for this to even have a real value, 4-4a must be positive or 0, so 4-4a>=0 means 4>=4a, 1>=a or a<=1.
ok....I'm going for it. hope you understand some of what I'm writing.
When the denominator is 0, this graph has asymptotes because fraction is undefined. x^2+4x+3a=0 when x=(-4+-sqrt 16-12a)/2 which is defined when 16-12a >=0 or a<=4/3.
So "a"must satisfy both so a<=1 which means this graph must have 2 asymptotes if it achieves all real numbers.
Huge simplification, asymptotes guarantee that the fraction achieves positive and negative infinity...sometimes, but doesn't guarantee it reaches all real numbers. your equation is guaranteed to cross the horizontal asymptote of y=1 when x=-a.......It's too much to write and I'm not sure you'd understand AND I'm Not sure I'm even answering what your teacher is trying to ask. Hope this was somewhat helpful.
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08/10/14
Shivam D.
08/09/14