y=0 is correct answer but is not in the selection.
Step-by-step explanation:
SOLUTION
For these solutions, we will use \displaystyle f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}, q\left(x\right)\ne 0f(x)=
q(x)
p(x)
,q(x)≠0.
\displaystyle g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}g(x)=
2x
3
+5x
2
6x
3
−10x
: The degree of \displaystyle p=\text{degree of } q=3p=degree of q=3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at \displaystyle y=\frac{6}{2}y=
2
6
or \displaystyle y=3y=3.
\displaystyle h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}h(x)=
x+2
x
2
−4x+1
: The degree of \displaystyle p=2p=2 and degree of \displaystyle q=1q=1. Since \displaystyle p>qp>q by 1, there is a slant asymptote found at \displaystyle \frac{{x}^{2}-4x+1}{x+2}
x+2
x
2
−4x+1
.
Synthetic division of the polynomial x^2-4x+1 by x+2 in which it only contains the coefficients of each polynomial.
The quotient is \displaystyle x - 6x−6 and the remainder is 13. There is a slant asymptote at \displaystyle y=x - 6y=x−6.
\displaystyle k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}k(x)=
x
3
���8
x
2
+4x
: The degree of \displaystyle p=2\text{ }<p=2 < degree of \displaystyle q=3q=3, so there is a horizontal asymptote y = 0.
