attempt 1: 2 attempts remaining. consider the function f(x)=(5x - 2)/((-2x - 8)(8x - 2)). enter the…

attempt 1: 2 attempts remaining. consider the function f(x)=(5x - 2)/((-2x - 8)(8x - 2)). enter the equations of the vertical asymptotes. if there are no vertical asymptotes, enter none. if there is more than one vertical asymptote, enter a list of the equations separated by a comma (e.g., x=20, x=-7). vertical asymptotes: / enter the equations of the horizontal asymptotes. if there are no horizontal asymptotes, enter none. if there is more than one horizontal asymptote, enter a list of the equations separated by a comma (e.g., y=20, y=-7). horizontal asymptotes: / submit answer next item

attempt 1: 2 attempts remaining. consider the function f(x)=(5x - 2)/((-2x - 8)(8x - 2)). enter the equations of the vertical asymptotes. if there are no vertical asymptotes, enter none. if there is more than one vertical asymptote, enter a list of the equations separated by a comma (e.g., x=20, x=-7). vertical asymptotes: / enter the equations of the horizontal asymptotes. if there are no horizontal asymptotes, enter none. if there is more than one horizontal asymptote, enter a list of the equations separated by a comma (e.g., y=20, y=-7). horizontal asymptotes: / submit answer next item

Answer

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to 0: $(-2x - 8)(8x - 2)=0$. Then solve $-2x - 8 = 0$ and $8x - 2=0$. For $-2x - 8 = 0$, we have $-2x=8$, so $x=-4$. For $8x - 2 = 0$, we have $8x=2$, so $x=\frac{1}{4}$.

Step2: Find horizontal asymptotes

The degree of the numerator is 1 and the degree of the denominator is 2 (since it's a product of two linear - functions). When the degree of the numerator $n = 1$ and the degree of the denominator $m = 2$ with $n<m$, the horizontal asymptote is $y = 0$.

Answer:

Vertical asymptotes: $x=-4,x=\frac{1}{4}$ Horizontal asymptotes: $y = 0$