For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.
b)
f(x)=+squareroot x−2
f(x) = √(x-2)
You know that the domain of √x is x>=0.
So, you need x-2 >= 0
Not too hard, eh?
To determine the domain of the function f(x) = √(x - 2), we need to consider the values of x for which the function is defined. In this case, since we have a square root function, the expression inside the square root must be non-negative. Thus, x - 2 ≥ 0. Solving this inequality, we find x ≥ 2. Therefore, the domain of f(x) is all real numbers greater than or equal to 2.
Next, let's find the range of the function. The range refers to the set of all possible output values of the function. Since we have a square root function, the range consists of all non-negative real numbers: y ≥ 0.
Regarding the restrictions, in this case, there are no explicit restrictions on the function.
To determine the intervals of increasing and decreasing, we need to consider the slope of the function. In this case, since we are dealing with a square root function, its graph starts at the point (2, 0) and continues to increase as x increases. Therefore, the function f(x) = √(x - 2) is increasing for all x ≥ 2.
To find the roots, we set the function equal to zero and solve for x:
√(x - 2) = 0
x - 2 = 0
x = 2
The only root of the function is x = 2.
To find the y-intercept, we substitute x = 0 into the function:
f(0) = √(0 - 2)
f(0) = √(-2)
Since the square root of a negative number is undefined in the real number system, there is no y-intercept for this function.
Lastly, to find the vertex of the function, we note that for a square root function of the form f(x) = √(x - h) + k, the vertex is at the point (h, k). Therefore, for f(x) = √(x - 2), the vertex is at (2, 0).
In summary:
Domain: x ≥ 2
Range: y ≥ 0
Restrictions: None
Intervals of increasing: x ≥ 2
Intervals of decreasing: None
Roots: x = 2
Y-intercept: None
Vertex: (2, 0)