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so i need to find the equation of these hyperbolas
i got the first one right, y = 3/x
but I don't know where I'm going wrong with the second one, i wrote y=4/(x-2) and i just don't get where I'm going wrong with it.

and i just can't figure out how to do number 2, 3 and 5, I don't know where to start (the one above 4 is 3)

please tell me how to get the equations​

so i need to find the equation of these hyperbolasi got the first one right y 3xbut I dont know where Im going wrong with the second one i wrote y4x2 and i just class=


Answer :

semsee45

Answer:

[tex]\textsf{2)}\quad y=\dfrac{5}{x}+3[/tex]

[tex]\textsf{3)}\quad y=-\dfrac{3}{x+2}[/tex]

[tex]\textsf{5)}\quad y=-\dfrac{3}{x+4}-3[/tex]

Step-by-step explanation:

A rectangular hyperbola is a specific type of hyperbola characterized by its asymptotes being perpendicular to each other, forming a right angle.

As the vertical asymptotes of the graphed hyperbolas are parallel to the y-axis and the horizontal asymptotes are parallel to the x-axis, they are perpendicular to each, indicating that the graphed hyperbolas are rectangular hyperbolas.

The general formula of a rectangular hyperbola with asymptotes that are parallel to the coordinate axes is:

[tex]y=\dfrac{a}{x-h}+k[/tex]

where:

  • (h, k) is the center of the hyperbola.
  • x = h is the vertical asymptote.
  • y = k is the horizontal asymptote.

[tex]\dotfill[/tex]

Question 2

From observation of the given hyperbola:

  • Vertical asymptote: x = 0
  • Horizontal asymptote: y = 3

Therefore, substitute h = 0 and k = 3 into the formula:

[tex]y=\dfrac{a}{x-0}+3 \\\\\\ y=\dfrac{a}{x}+3[/tex]

To find the value of a, substitute a point on the curve. Let's use (5, 4):

[tex]4=\dfrac{a}{5}+3 \\\\\\ 1=\dfrac{a}{5} \\\\\\ a=5[/tex]

Therefore, the equation of the graphed hyperbola is:

[tex]y=\dfrac{5}{x}+3[/tex]

[tex]\dotfill[/tex]

Question 3

From observation of the given hyperbola:

  • Vertical asymptote: x = -2
  • Horizontal asymptote: y = 0

Therefore, substitute h = -2 and k = 0 into the formula:

[tex]y=\dfrac{a}{x-(-2)}+0 \\\\\\ y=\dfrac{a}{x+2}[/tex]

To find the value of a, substitute a point on the curve. Let's use (-3, 3):

[tex]3=\dfrac{a}{-3+2} \\\\\\ 3=\dfrac{a}{-1} \\\\\\ a=-3[/tex]

Therefore, the equation of the graphed hyperbola is:

[tex]y=-\dfrac{3}{x+2}[/tex]

[tex]\dotfill[/tex]

Question 5

From observation of the given hyperbola:

  • Vertical asymptote: x = -4
  • Horizontal asymptote: y = -3

Therefore, substitute h = -4 and k = -3 into the formula:

[tex]y=\dfrac{a}{x-(-4)}-3 \\\\\\ y=\dfrac{a}{x+4}-3[/tex]

To find the value of a, substitute a point on the curve. Let's use (-5, 0):

[tex]0=\dfrac{a}{-5+4}-3 \\\\\\ 3=\dfrac{a}{-1} \\\\\\a=-3[/tex]

Therefore, the equation of the graphed hyperbola is:

[tex]y=-\dfrac{3}{x+4}-3[/tex]

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