How do you find the vertical, horizontal and slant asymptotes of #b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)#?

Question

Elijah Allen · Accepted Answer

Vertical asymptotes : # uarr x = 0.861 darr and uarr x = -1.227 darr#

The vertical asymptotes are given by

x = zero of #d(x)=2x^4+x^3-x^2-1#

By considering the chanes 1 and 1 in the signs of the coefficients of

d(x), d has either two real roots are none,

The first graph for d reveals two real zeros, near 0.8 and -1.2.

The second near- linear mapping near x = 0,8 reveals this zero as

0.861, nearly, and likewise, the third graph reveals the second zero

near-1.227.

graph{2x^4+x^3-x^2-1 [-1.25, 1.25, -0.625, 0.625]}

graph{2x^4+x^3-x^2-1 [.86, .87, -.43.43]}

graph{2x^4+x^3-x^2-1 [-1.228, -1.226, -.43.43]}

Anna Allen · Answer

undefined To find the vertical, horizontal, and slant asymptotes of the function $b(x) = \frac{x^3 - x + 1}{2x^4 + x^3 - x^2 - 1}$, we analyze its behavior as $x$ approaches infinity and negative infinity, as well as any potential points of discontinuity.

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of the function equals zero, but the numerator does not. To find them, set the denominator equal to zero and solve for $x$.

$$2x^4 + x^3 - x^2 - 1 = 0$$

Factoring, we get:

$$ (2x^2 - x + 1)(x^2 + 1) = 0$$

The quadratic $2x^2 - x + 1$ does not have real roots, so the only factor that matters is $x^2 + 1 = 0$, which has complex roots. Therefore, there are no vertical asymptotes.

2. **Horizontal Asymptotes**: Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. In this case, the degree of the numerator is 3 and the degree of the denominator is 4, so there is a horizontal asymptote at $y = 0$, which is the x-axis.

3. **Slant Asymptotes**: Slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator. To find the slant asymptote, perform polynomial long division or use a similar method to divide the numerator by the denominator.

$$ \frac{x^3 - x + 1}{2x^4 + x^3 - x^2 - 1} = \frac{1}{2}x^{-1} - \frac{3}{4}x^{-2} + \frac{3}{8}x^{-3} - \frac{15}{16}x^{-4} + R(x)$$

As $x$ approaches infinity, the remainder term $R(x)$ approaches zero, leaving us with the slant asymptote:

$y = \frac{1}{2}x^{-1} - \frac{3}{4}x^{-2} + \frac{3}{8}x^{-3} - \frac{15}{16}x^{-4}$

In summary, the function $b(x)$ has no vertical asymptotes, a horizontal asymptote at $y = 0$, and a slant asymptote at $y = \frac{1}{2}x^{-1} - \frac{3}{4}x^{-2} + \frac{3}{8}x^{-3} - \frac{15}{16}x^{-4}$.