How do you find the vertical, horizontal and slant asymptotes of: #y=(8x-48)/(x^2-13x+42)#?

Question

Lincoln Anderson · Accepted Answer

vertical asymptote x = 7
horizontal asymptote y = 0

The first step here is to factorise and simplify y. #(8x-48)/(x^2-13x+42)=(8cancel((x-6)))/((x-7)cancel((x-6)))=8/(x-7)# The denominator of this rational function cannot be zero as this would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the value that x cannot be and if the numerator is also non-zero for this value of x then it must be a vertical asymptote. solve : x - 7 = 0 → x = 7 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),ytoc" (a constant)"# divide terms on numerator/denominator by x #(8/x)/(x/x-7/x)=(8/x)/(1-7/x)# as #xto+-oo.yto0/(1-0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0 ,denominator-degree 1 )Hence there are no slant asymptotes. graph{8/(x-7) [-20, 20, -10, 10]}

Greyson Brown · Answer

undefined To find the vertical asymptotes, we need to determine the values of $x$ that make the denominator of the rational function equal to zero. These values are the vertical asymptotes.

To find the horizontal asymptote, we examine the degrees of the numerator and denominator polynomials. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degrees are equal, we take the ratio of the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

To find the slant asymptote, we perform polynomial long division to divide the numerator by the denominator. The quotient obtained represents the equation of the slant asymptote.

Let's proceed with the calculations.

1. Vertical asymptotes:
Set the denominator equal to zero and solve for $x$.
$$x^2 - 13x + 42 = 0$$
$(x - 6)(x - 7) = 0$
$x = 6$ or $x = 7$

So, there are vertical asymptotes at $x = 6$ and $x = 7$.

2. Horizontal asymptote:
The degree of the numerator is 1, and the degree of the denominator is 2. So, there is a horizontal asymptote at $y = 0$.

3. Slant asymptote:
Perform polynomial long division:
$$ \frac{8x - 48}{x^2 - 13x + 42} = \frac{8x - 48}{(x - 6)(x - 7)} $$
$$ = \frac{8(x - 6)}{(x - 6)(x - 7)} $$
$$ = \frac{8}{x - 7} $$

So, the slant asymptote is $y = \frac{8}{x - 7}$.