can someone help me find the domain and range of the function f(x)=3(x-3)^2-1
I think the domain is all real numbers but I'm not sure about the range
nevermind I figured it out
good work. Just for the future, recall that the domain of all polynomials is all real numbers.
To find the domain and range of the function f(x) = 3(x-3)^2 - 1, it's important to understand what these terms mean.
The domain of a function refers to all the possible input values (x-values) that the function can accept. In other words, it represents the set of values for which the function is defined. To determine the domain, consider any potential restrictions on x.
In this case, there are no specific restrictions or exclusions to the function f(x) = 3(x-3)^2 - 1. Therefore, the domain is all real numbers, or (-∞, ∞).
The range of a function, on the other hand, refers to all the possible output values (y-values) that the function can produce. It represents the set of values that the function can take on. To determine the range, you can utilize a variety of methods.
One way to find the range is by analyzing the graph of the function. However, given the function f(x) = 3(x-3)^2 - 1, we can also determine the range without graphing.
Since the function is in the form f(x) = a(x-h)^2 + k, where a, h, and k are constants, we know that the graph of the function is a parabola that opens upwards if a > 0.
In this case, a = 3, which is positive, so the parabola opens upwards.
To find the range, we need to find the minimum value of the function. In this case, the minimum value occurs when the vertex of the parabola is reached. The x-coordinate of the vertex is given by h = 3, and the y-coordinate is given by k = -1.
Therefore, the minimum value of f(x) is -1, and since the parabola opens upwards, the range of the function is all values y ≥ -1. In interval notation, the range is represented as [-1, ∞).
In summary:
Domain: All real numbers (-∞, ∞)
Range: All values y ≥ -1 ([-1, ∞))