Create a function with the given features: a vertical asymptote the y-axis, an oblique asymptote y=3x+1, and no x- and y-intercepts. Please explain.First of all, the answer is f(x) = 3x^2 + x + 5...

Create a function with the given features: a vertical asymptote the y-axis, an oblique asymptote y=3x+1, and no x- and y-intercepts. Please explain.

First of all, the answer is f(x) = 3x^2 + x + 5 divided by or over x. I can get part of the answer as I can only get the denominator, but I don't understand how a 5 ends up in the question. The answer also states that the constant in the numerator could be  any real number provided the discriminant of the quadratic numerator is <0, therefore ensuring that the rational function has no x-intercept. Thanks in advance if your are able to help me understand this.

Asked on by islnds

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neela | High School Teacher | (Level 3) Valedictorian

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We know that f(x) and g(x) are asymptotic if f(x)/g(x)  = 1 as x--> infinity. Or f(x) - g(x) = k a constant as x --> infinity.

We consider the function y(x) = (3x^2+5x+5)/x

Let f(x) = 3x+5.

Then y(x) - 3x+5 = (3x^2+5x+5)/x - 3x+5  = {3x^2+5x+5-3x^2-5x) = 5/x  > 0 x for all x ant 5/x = 0 as x --> infinity.

Therefore  3x+1 is the oblique asymptote.

Also  for a vertical asymptote, at  x= a , y should be infinite.

So at x = 0,  y(x) = (3x^2+x+5)/x =  3x+1 +5/x  is infinte as 5/x = ifinity as x--> 0.

Threfore 3x^2+1  and y axis  are the asmptotes for the curve y = (3x^2+x+5)/x.

 

 

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