Rational function 4x^4-2x+3/2x^3-6x^2+1

Question

4x^4-2x+3/2x^3-6x^2+1

I need to find the vertical asymptote, horizonal asymptote or slant asymptote. I graphed it using a online graphing tool and it looks really weird, cannot figure out the asymptotes. Since the numerator is exactly one degree higher than denominator, i used poly long division and got 2x+8, so this is the slant asymptote? I used the rational zero theorem calculator to find x intercept for the denominator, but there was none.

Answer 1

1. Your slant asymptote is almost right. Perform more carefully the long division.
2. The equation 2x³-6x²+1=0 indeed does not have rational roots. But as a third degree polynomial

has at least one real root . Therefore graph the function F(x)= 2x³-6x²+1 and the zero's of this

function are the vertical asymptotes of the original one. In our case there are three not rational

numbers that make the denominator zero

Answer 2

The degree of the numerator is greater than the degree of the denominator, so there are no horizontal asymptotes. You can see this from the plot of the rational function.

For the vertical asymptotes, need to set the denominator to zero and solve for x. There is no "nice" cubic factoring for this denominator, so have to use the cubic formula to find the 3 values of x for the vertical

asymptotes.

I got x=−0.384367,0.442125,2.942242. If you add these vertical lines to your plot, you will see that these are the vertical asymptotes.

Since the degree of the numerator is greater than the denominator, for the slant asymptote, from long division, I got 2x+6+Remainder, so 2x+6 is the slant asymptote. If you plot y = 2x+6 against the rational function, you will see that this line is the slant asymptote.