Find the domain and range of the function y= -3(x-1/3)^2+4/3
as with all polynomials, the domain is (-∞,∞)
Since the graph is a parabola, it has vertex.
The leading coefficient is negative, so the parabola opens downward.
So, the range is everything which is below the vertex: (-∞, 4/3]
Your equation is in the usual vertex form.
You can read off the vertex to be (1/3 , 4/3)
You also know that the parabola opens downwards (from the -3),
so your vertex is a maximum.
Of course the domain of such a parabola is the set of real numbers.
Tell me what you think about the range.
@oobleck I don't understand
@Reiny the range would be 4/3?
The range would be :
y ≤ 4/3 , y ∊ R
or as oobleck stated, (-∞, 4/3]
To find the domain and range of the function y = -3(x - 1/3)^2 + 4/3, we need to consider two aspects: the domain and the range.
1. Domain:
The domain refers to all the possible values that x can take for the given function. In this case, there are no restrictions or limitations on the variable x since it is not divided by zero or involved in any other operations that could be undefined. Thus, the domain is all real numbers (-∞, +∞).
2. Range:
The range represents all the possible values that y can take for the given function. Let's analyze the function to determine the range. We notice that it is a quadratic function in the form y = ax^2 + bx + c, where a = -3, b = 2/3, and c = 4/3.
Since the coefficient "a" in this quadratic function is negative, the parabola will open downwards. The vertex of the parabola occurs at the value of x = -b/2a, which we can calculate as follows:
x = -b/2a
= -(2/3) / (2 * -3)
= -2/3 / -6
= -2/3 * -1/6
= 2/18
= 1/9
Therefore, the vertex occurs at (1/9, 4/3).
Knowing the vertex, we can conclude that the vertex represents the minimum point of the parabola and y can have any value greater than or equal to the y-coordinate of the vertex. Hence, the range is [4/3, +∞).
Summary:
- Domain: (-∞, +∞)
- Range: [4/3, +∞)