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Sagot :
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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