anonymous
  • anonymous
What are the equations of the asymptotes of the rational function?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
It's gotta be the third option because that's the only one that has the vertical asymptote at x=7 ; which is really easy to spot looking at the denominator of the function and determining what value causes division by zero.
anonymous
  • anonymous
I'm sorry - I didn't mean to use the word "easy".

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anonymous
  • anonymous
Vertical asymptotes are generally quicker to spot than horizontal asymptotes or slant asymptotes because students tend to be programmed to determine which values to immediately throw out of the domain of a function like that.
Directrix
  • Directrix
Looking at the equation, y = 5/(x -7) + 6, note that the denominator of the first section of the equation is x - 7. If x were to equal 7, then 7 -7 = 0 which would result in 5/0. Division by zero is undefined so x cannot be 7 in the function. The vertical asymptote with equation x = 7 is obtained. The graph cannot cross its vertical asymptote. For the horizontal asymptote, consider what happens to 5/(x -7) + 6 as x gets larger and larger (increases without bound). In a fraction with a contant numerator and a larger and larger denominator, the values of the fraction become closer and closer to 0. The contant term 6 is not affected by the values of x. Therefore, the horizonal asymptote is y = 6. If you have studied limits, you can take the limit of f(x) as x --> ∞ and get the same result. It's always nice to reinforce the algebraic concepts with visual depictions. I'll try to get a graph that shows the function.

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