In general for the function y = √x
For the square root function, only positive values are possible values of x. For negative values, the function is undefined. So,
From the function finding domain and range :
Fix the radicand ≥ 0 and solve for x.
D is [0, ∞) R is [0, ∞) | |
Reflection across x-axis, D is [0, ∞) R is (-∞, 0] | |
Reflection across y-axis, D is (-∞, 0] R is [0, ∞) |
By finding horizontal asymptote, we can fix the starting point simply. From there we can fix the range.
For the function which is in the form
y = a√b(x-h) + k
y = k is the horizontal asymptote.
Describe the domain of the function.
Problem 1 :
y = 8√x
Solution :
There is no reflection.
Domain :
√x ≥ 0
x ≥ 0
Domain is [0, ∞)
Range :
Horizontal asymptote is y = 0. The range is [0, ∞).
Problem 2 :
y = 4 + √-x
Solution :
There is reflection across y-axis.
Domain :
√-x ≥ 0
-x ≥ 0
x ≤ 0
Domain is (-∞, 0].
Range :
Horizontal asymptote is y = 4. The range is [4, ∞).
Problem 3 :
y = -√(x - 3)
Solution :
There is reflection across x-axis.
Domain :
√(x - 3) ≥ 0
x - 3 ≥ 0
x ≥ 3
Domain is [3, ∞).
Range :
Horizontal asymptote is y = 0. The range is (-∞, 0]
Problem 4 :
y = √(x + 2) - 2
Solution :
There is no reflection.
Domain :
√(x + 2) ≥ 0
x + 2 ≥ 0
x ≥ -2
Domain is [-2, ∞).
Range :
Horizontal asymptote is y = -2. The range is [-2, ∞).
Problem 5 :
y = √(-x - 1)
Solution :
y = √(-x - 1)
y = √-(x + 1)
There is reflection across y-axis.
Domain :
√-(x + 1) ≥ 0
-(x + 1) ≥ 0
(x + 1) ≤ 0
x ≤ -1
Domain is (-∞, -1].
Range :
Horizontal asymptote is y = 0. The range is [0, ∞).
Problem 6 :
y = √(2x) + 7
Solution :
There is no reflection
Domain :
√2x ≥ 0
2x ≥ 0
Dividing by 2 on both sides
x ≥ 0
Domain is [0, ∞).
Range :
Horizontal asymptote is y = 7. The range is [7, ∞).
Problem 7 :
y = (1/2)√(-x - 2)
Solution :
y = (1/2)√(-x - 2)
y = (1/2)√-(x + 2)
There is reflection across y-axis.
Domain :
√-(x + 2) ≥ 0
(x + 2) ≤ 0
x ≤ -2
Domain is (-∞, -2].
Range :
Horizontal asymptote is y = 0. The range is [0, ∞).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM