anonymous
  • anonymous
What is the equation of the following graph?
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.
chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
that is a traslated hyperbola

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anonymous
  • anonymous
|dw:1439232785706:dw| I have to use this equation right?
Michele_Laino
  • Michele_Laino
the requested equation is an equation like this: \[\Large \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{b^2}}} = 1\]
Michele_Laino
  • Michele_Laino
from your graph we can see that: \[\Large 2a = 10\]
anonymous
  • anonymous
wait, how do you know A is 2?
anonymous
  • anonymous
isn't a "x" and b "y" ?
Michele_Laino
  • Michele_Laino
what is A?
Michele_Laino
  • Michele_Laino
referring to your drawing, we have: 2a=10, to understand that it is suffice to count the number of little squares
Michele_Laino
  • Michele_Laino
namely 2*a is the base of the dashed rectangle, whereas 2*b is the height of the dashed rectangle so 2*b=4
Michele_Laino
  • Michele_Laino
then we have: \[\Large \begin{gathered} \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{5^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{2^2}}} = 1 \hfill \\ \hfill \\ \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{4} = 1 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now, we establish what are x_0 and y_0
anonymous
  • anonymous
im so confused ...
Michele_Laino
  • Michele_Laino
from your graph, I see that your hyperbola passes at point (6,2), so we have: \[\Large \frac{{{{\left( {6 - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - {y_0}} \right)}^2}}}{4} = 1\]
anonymous
  • anonymous
but how did you get the 25?
Michele_Laino
  • Michele_Laino
here is why:
Michele_Laino
  • Michele_Laino
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anonymous
  • anonymous
Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?
Michele_Laino
  • Michele_Laino
as I said before, we have to establish the value of x_0 and y_0
anonymous
  • anonymous
and how do we do that?
Michele_Laino
  • Michele_Laino
as we can see, the center of our dashed rectangle is located at point (1,2), right?
Michele_Laino
  • Michele_Laino
do you see that point?
anonymous
  • anonymous
right!
Michele_Laino
  • Michele_Laino
in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up
anonymous
  • anonymous
okay, and what is the next step now?
Michele_Laino
  • Michele_Laino
that means x_0=1 and y_0=2, so, substituting them intoi my equation above, we get: \[\Large \frac{{{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1\]
Michele_Laino
  • Michele_Laino
into*
anonymous
  • anonymous
so where did they 6 & 2 go?
Michele_Laino
  • Michele_Laino
x=6 and y=2 are the coordinates of a point which belongs to our hyperbola, infact, if we substitute x=6 and y=2 into my last formula, we get: \[\large {\text{left side}} = \frac{{{{\left( {6 - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - 2} \right)}^2}}}{4} = \frac{{25}}{{25}} - 0 = 1 = {\text{right side}}\] namely I got an identity

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