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Find the vertical and horizontal asymptotes of the function:

f(x)=((2x-1)(3x+1))/((x-2)(x+4))
The fields below accept a list of numbers or formulas separated by semicolons (e.g.
2;4;6 or
x+1;x-1). The order of the list does not matter.
Vertical asymptotes:

x=
Horizontal asymptotes:

y=

◻

$\newline$ Find the vertical and horizontal asymptotes of the function: $\newline$ $f(x)=\frac{(2 x-1)(3 x+1)}{(x-2)(x+4)}$ $\newline$ The fields below accept a list of numbers or formulas separated by semicolons (e.g. $2 ; 4 ; 6$ or $x+1 ; x-1)$ . The order of the list does not matter. $\newline$ Vertical asymptotes: $\newline$ $x=$ $\newline$ Horizontal asymptotes: $\newline$ $y=$ $\newline$

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Full solution

Q.

\newline

Find the vertical and horizontal asymptotes of the function:

\newline

f(x)=\frac{(2 x-1)(3 x+1)}{(x-2)(x+4)}

\newline

The fields below accept a list of numbers or formulas separated by semicolons (e.g.

2 ; 4 ; 6

or

x+1 ; x-1)

. The order of the list does not matter.

\newline

Vertical asymptotes:

\newline

x=

\newline

Horizontal asymptotes:

\newline

y=

\newline

Find Vertical Asymptotes: To find the vertical asymptotes, set the denominator equal to zero and solve for $x$ . $(x-2)(x+4) = 0$ $x-2 = 0 \text{ or } x+4 = 0$ $x = 2 \text{ or } x = -4$
Compare Degrees for Horizontal Asymptotes: For the horizontal asymptotes, compare the degrees of the numerator and the denominator. $\newline$ Degree of numerator $(2x-1)(3x+1) = 2$ (since it's a product of two first-degree polynomials). $\newline$ Degree of denominator $(x-2)(x+4) = 2$ (same reason as above).
Calculate Horizontal Asymptote: Since the degrees are the same, the horizontal asymptote is found by dividing the leading coefficients of the numerator and the denominator. $\newline$ Leading coefficient of numerator = $2 \times 3 = 6$ $\newline$ Leading coefficient of denominator = $1 \times 1 = 1$ $\newline$ Horizontal asymptote: $y = \frac{6}{1} = 6$

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