#### Graphs of the hyperbola - answers.

2
3
1.1
y = ––––––– + 3
1.2
y = ————  – 4
x + 1
x + 2
Horizontal asymptote: y = 3
Horizontal asymptote: y = −4
Vertical asymptote: x = − 1
Vertical asymptote: x = − 2
2
3
X-intercept: ––––––– + 3 = 0
X-intercept: –––––––  − 4 = 0
x + 1
x + 2
X (x + 1) :     2 + 3(x + 1) = 0
X (x + 2) :     3 - 4(x + 2) = 0
2 + 3x + 3 = 0
3 - 4x − 8 = 0
x = − 1,67
x = − 1,2
X-intercept is (− 1,67 ; 0)
X-intercept is (− 1,2 ; 0)
2
3
Y-intercept: y = ––––––– + 3
Y-intercept: y = –––––––  − 4
0 + 1
0 + 2
= 5
= − 2,5
Y-intercept is (0 ; 5)
Y-intercept is (0 ; − 2,5)
a is positive and thus the graph is in quadrants I and III
a is positive and thus the graph is in quadrants I and III
5
− 2
1.3
y = ––––––– + 5
1.4
y = ————  + 3
x - 3
x + 1
Horizontal asymptote: y = 5
Horizontal asymptote: y = 3
Vertical asymptote: x = 3
Vertical asymptote: x = − 1
5
− 2
X-intercept: ––––––– + 5 = 0
X-intercept: –––––––  + 3 = 0
x - 3
x + 1
X (x - 3) :     5 + 5(x - 3) = 0
X (x + 1) :     −2 + 3(x + 1) = 0
5 + 5x - 15 = 0
− 2 + 3x + 3 = 0
x = 2
x = − 0,33
X-intercept is (2 ; 0)
X-intercept is (− 0,33 ; 0)
5
− 2
Y-intercept: y = ––––––– + 5
Y-intercept: y = –––––––  + 3
0 - 3
0 + 1
= 3,33
= 1
Y-intercept is (0 ; 3,33)
Y-intercept is (0 ; 1)
a is positive and thus the graph is in quadrants I and III
a is negative and thus the graph is in quadrants II and IV
− 3
− 5
1.5
y = –––––––  − 4
1.6
y = ————  − 2
x − 2
x + 3
Horizontal asymptote: y = − 4
Horizontal asymptote: y = − 2
Vertical asymptote: x = 2
Vertical asymptote: x = − 3
− 3
− 5
X-intercept: –––––––  − 4 = 0
X-intercept: –––––––  − 2 = 0
x − 2
x + 3
X (x − 2) :    −3 − 4(x − 2) = 0
X (x + 3) :    −5 − 2(x + 3) = 0
−3 − 4x + 8 = 0
− 5 − 2x − 6 = 0
x = 1,25
x = − 5,5
X-intercept is (1,25 ; 0)
X-intercept is (− 5,5 ; 0)
− 3
− 5
Y-intercept: y = –––––––  − 4
Y-intercept: y = –––––––  − 2
0 - 2
0 + 3
= − 2,5
= − 3,67
Y-intercept is (0 ; − 2,5)
Y-intercept is (0 ; − 3,67)
a is negative and thus the graph is in quadrants II and IV
a is negative and thus the graph is in quadrants II and IV
2.1
Horizontal asymptote: y = 3 and thus q = 3
2.2
Horizontal asymptote: y = − 3 and thus q = −3
Vertical asymptote: x = −1 and thus p = 1
Vertical asymptote: x = 1 and thus p = −1
a
a
Equation is y = –––––––  + 3
Equation is y = –––––––  − 3
x + 1
x − 1
Now substitute the coordinates of a point for x and y in the equation and solve for a.
a
a
At A(−2 ; 0) :  –––––––  + 3 = 0
At B(0 ; −8) :  –––––––  − 3 = −8
−2 + 1
0 − 1
a
a
–––––––  + 3 = 0
–––––––  − 3 = −8
−1
−1
X −1 :                   a − 3 = 0
X −1 :                    a + 3 = 8
a = 3
a = 5
3
5
Equation is y = –––––––  + 3
Equation is y = –––––––  − 3
x + 1
x − 1
2.3
Horizontal asymptote: y = 4 and thus q = 4
2.4
Horizontal asymptote: y = − 1 and thus q = −1
Vertical asymptote: x = −2 and thus p = 2
Vertical asymptote: x = 3 and thus p = −3
a
a
Equation is y = –––––––  + 4
Equation is y = –––––––  − 1
x + 2
x − 3
Now substitute the coordinates of a point for x and y in the equation and solve for a.
a
a
At Q(−3 ; 7) :  –––––––  + 4 = 7
At Q(5 ; −3) :  –––––––  − 1 = −3
−3 + 2
5 − 3
a
a
–––––––  + 4 = 7
–––––––  − 1 = −3
−1
2
X −1 :                   a − 4 = − 7
X 2 :                    a − 2 = − 6
a = − 3
a = − 4
− 3
− 4
Equation is y = –––––––  + 3
Equation is y = –––––––  − 3
x + 2
x − 3
3.1
P is the point (1 ; 2)
3.2
p = − 1 and q = 2
a
3.3
y = (x + p) + q
Equation is y = –––––––  + 2
x − 1
a
y = (x − 1) + 2
At Q(2;5) :  –––––––  + 2 = 5
2 − 1
a
y = x − 1 + 2
–––––––  + 2 = 5
1
y = x + 1
a + 2 = 5
a = 3
3
Equation is y = –––––––  + 2
x − 1
3.4
At R the equations are equal so that
3
–––––––   + 2 = x + 1
x − 1
X (x − 1)  :      3 + 2x − 2 = x2 − 1
x2 − 2x − 2 = 0
———————
−(−2) ± √(−2)2 −4(1)(−2)
x =  ————————————
2(1)
x = 2,73  OR  x = − 0,73
At R x > 0 and thus x = 2,73 and y = 2,73 + 1 = 3,73
R is the point (2,73 ; 3,73)
4.1
P is the point (−2 ; −3) and therefore p = 2 and q = − 3
a
5
4.2
Equation is y = –––––––  + 2
4.3
At B : y = –––––––  − 3
x + 2
0 + 2
a
5
At Q(−1 ; 2):  –––––––  − 2 = 2
= — − 3
− 1 + 2
2
a − 3  = 2
= − 0,5
5
a = 5
At A :   –––––––  − 3 = 0
x + 2
4.4
y = (x + p) + q
X (x + 2):   5 − 3x − 6 = 0
y = (x + 2) − 3
x = − 0,33
= x − 1
5
4.6
Domain : x = {x | x ≠ − 2 ; }
}
4.5
At T      –––––––  − 3 = x − 1
x + 2
Range : x = {y | y ≠ − 3 ; }
}
X (x + 2):   5 − 3x − 6 = x2 + x − 2
x2 + 4x − 3 = 0
—————
−4 ± √42 −4(1)(−3)
x =  ——————————
x = − 4,65  OR  x = 0,65
5.1
Horizontal asymptote: y = 4 and thus q = 4
5.2
Vertical asymptote: x = 2 and thus p = − 2
a
− 8
5.3
Equation is y = –––––––  + 4
5.4
At A :      –––––––  + 4 = 0
x − 2
x − 2
a
At Q(−2;6) :     –––––––  + 4 = 6
X (x − 2):    −8 + 4x −8 = 0
−2 −2
X −4:                           a −16 = −24
4x = 16
a = −8
x = 4
− 8
− 8
Equation is y = –––––––  + 4
At B : y =   –––––––  + 4
x − 2
0 − 2
= 4 + 4 = 8
A is the point (4 ; 0) and B is the point (0 ; 8)
5.5
Equation is y = −( x + p ) + q
5.6
At R and S the y-values are equal.
− 8
y = −( x − 2 ) + 4
–––––––  + 4 = − x + 6
x − 2
= − x + 2  + 4
X ( x − 2 ) :  − 8 + 4(x − 2) = (x − 2)(−x + 6)
= − x + 6
− 8 + 4x − 8 = − x2 + 8x − 12
x2 − 4x − 4 = 0
———————
−(−4) ± √(−4)2 −4(1)(−4)
5.7
Domain :  {x | x ≠ − 2 ; }
}
x =  ————————————
2(1)
—–
4 ± √32
Range :  {y | y ≠ 4 ; }
}
x =  —————
2
− 8
5.8
h(x) =   ––––––––––  + 4 − 3
x = − 0,83  OR  x = 4,83
x + 2 − 5
− 8
=   –––––––  + 1
y = 6,83  OR  y = 1,17
x − 3
6.1
Horizontal asymptote: y = 5 and thus q = 5
6.2
Vertical asymptote: x = 3 and thus p = − 3
6
6
6.3
5 − –––––––  = 0
6.4
y = 5 − –––––––
x − 3
0 − 3
X (x − 3) : 5x − 15 − 6 = 0
= 5 − (− 2)
5x = 21
= 7
x = 4,2
A is the point (4,2 ; 0)
B is the point (0 ; 7)
6
6
6.5
At D(5;d):  d = 5 − –––––––
6.6
At E(e;−1): 5 − ––––––– = 6,5
5 − 3
e − 3
= 5 − 3
X(x − 3):  5e − 15 − 6 = 6,5e − 19,5
= 2
− 1,5e = 1,5
e = − 1
6.7
At R and S the y-values of the hyperbola and the straight line are equal.
6
At R and S:  5 − –––––––    = − 2x + 15
x − 3
X (x − 3):         5x − 15 − 6  = − 2x2 + 21x − 45
2x2 − 16x + 24 = 0
x2 − 8x + 12 = 0
(x − 2)(x − 6) = 0
x = 2  OR  x = 6
x = 2 : y = −2(2) + 15      OR      x = 6 : y = −2(6) + 15
= 11                   OR                    = 3
The points of intersection are (2 ; 11)     OR   (6 ; 3)
7.1
Horizontal asymptote: y = −3 and vertical asymptote: x = −1 and thus P is the point (−1 : −3)
− 5
− 5
7.2
–––––––   − 3 = 0
7.3
y =  –––––––   − 3
x + 1
0 + 1
X (x + 1) : − 5 − 3x − 3 = 0
= − 5 − 3
− 3x = 8
= − 8
x = − 2,67
A is the point (− 2,67 ; 0)
B is the point (0 ; − 8)
− 5
− 5
7.4
At D(0,5;d):  d = ––––––– − 3
7.5
At E(e;−1): –––––––  − 3  =  − 1
0,5 + 1
e + 1
− 19
=  –––––
X (e + 1) : − 5 − 3e − 3 = − e − 1
3
2 e = − 7
e = − 3,5
7.6
Positive axis of symmery: y = (x + p) + q    and  negative axis of symmetry: y = − (x + p) + q
y = (x + 1) − 3
y = − (x + 1) − 3
y = x − 2
y = − x − 4
4
4
8.1
Equation :  y = ––––––– + 2
Equation :  y = –––––––  − 2
x − 1
x + 2
Horizontal asymptote: y = 2
Horizontal asymptote: y = −2
Vertical asymptote: x = 1
Vertical asymptote: x = −2
Pieces are p and r
Pieces are q and 2
4
4
4
8.2
––––––– + 2  =    –––––––––––  + 2 − 4   =  ––––––––  − 2
x − 1
x − 1 + 3
x + 2
Graph A
Graph B
The graph A is translated 3 units to the left and 4 units downwards to form the graph B