How do you find the asymptotes for #y= (2x^3-5x+3) / (x^2-5x+4)#?

Question

Chloe Alexander · Accepted Answer

Vertical asymptote at #x=4# and slanting asymptote at #y=2x#

In #y=(2x^3-5x+3)/(x^2-5x+4)#, the factors of denominator are #(x-4)(x-1)# and numerator is also divisible by #x-1#, as such

#y=(2x^3-5x+3)/(x^2-5x+4)=((x-1)(2x^2+2x-3))/((x-4)(x-1))=(2x^2+2x-3)/(x-4)#

Hence, as denominator will be zero for #x=4#, we will have vertical asymptote at #x=4#.

Further as degree of numerator in #(2x^2+2x-3)/(x-4)# is greater than that of denominator and ratio of highest degrees by one in the two is #2x^2/x=2x#, we will have a slanting asymptote at #y=2x#

graph{(2x^3-5x+3)/(x^2-5x+4) [-15, 15, -5, 5]}

Benjamin Achtenberg · Answer

undefined To find the asymptotes of the given function $y = \frac{2x^3 - 5x + 3}{x^2 - 5x + 4}$, we first check for any vertical asymptotes by identifying the values of $x$ for which the denominator equals zero. Then, we look for horizontal asymptotes by examining the behavior of the function as $x$ approaches positive or negative infinity.

To find vertical asymptotes, we solve the equation $x^2 - 5x + 4 = 0$ for $x$. Factoring the quadratic expression, we have $(x - 1)(x - 4) = 0$, which gives us $x = 1$ and $x = 4$. Therefore, the vertical asymptotes occur at $x = 1$ and $x = 4$.

To find horizontal asymptotes, we examine the behavior of the function as $x$ approaches positive or negative infinity. We can use the concept of limits to analyze this behavior. As $x$ becomes very large in magnitude, the higher-degree terms in the numerator and denominator dominate the function. Since the degree of the numerator is greater than the degree of the denominator, the function does not have a horizontal asymptote.

In summary, the vertical asymptotes for $y = \frac{2x^3 - 5x + 3}{x^2 - 5x + 4}$ occur at $x = 1$ and $x = 4$, and there are no horizontal asymptotes.