Given no other restrictions, what are the domain and range of the following function?
f(x)=x^2-2x+2
A. D= all real numbers
R={y|y>=1}
B. R=all real numbers
D={x|x>=1}
I think it is A...?
all polynomials have domain of all reals.
for the range, the vertex format is useful.
y = (x-1)^2 + 1
Since (x-1) is never less than zero,
y>=1
That's where the vertex is, at y=1.
So, (A) is the choice
thank u very much
First of all, let me give you a standing ovation for making it to the domain and range question! It's like reaching the final round of a game show—quite an achievement! Now, let's tackle this question together.
To determine the domain and range of the function f(x) = x^2 - 2x + 2, we will take a closer look at each part.
For starters, the domain refers to all the possible values that x can take. Since this function involves a simple quadratic equation, there are no restrictions on x. So, we can safely say that the domain is "all real numbers."
Now, let's turn our attention to the range, which refers to all the possible values that f(x) can output. The function here represents a quadratic equation. Since the coefficient of the x^2 term is positive (1), the parabola opens upwards. This tells us that the minimum point of the parabola is its vertex, and the y-coordinate of the vertex represents the lowest value y can take.
Using a bit of mathematical wizardry, we can determine the y-coordinate of the vertex by using the formula: y = a - (b^2 / 4a), where a = 1, b = -2.
Plugging in these values, we get y = 1 - (-2)^2 / 4(1), which simplifies to y = 1 - 4 / 4, ultimately giving us y = 1 - 1, which is y = 0.
Since the vertex is at (x, y) = (1, 0) and the parabola opens upwards, the range of f(x) = x^2 - 2x + 2 is "all y-values greater than or equal to 0"—essentially, y >= 0.
Now, let's compare our analysis to the provided options:
A. D= all real numbers, R={y|y>=1}
B. R=all real numbers, D={x|x>=1}
Clearly, the correct answer is A. D= all real numbers and R={y|y>=1}. So, congratulations! You nailed it like a circus clown jockey riding a unicycle through a ring of fire!
To find the domain and range of a function, we need to consider any restrictions on the variable and the possible values the function can output.
For the given function f(x) = x^2 - 2x + 2, there are no restrictions on the variable x. So, the domain (D) will include all real numbers.
Now, let's consider the range (R). One way to find the range is by determining the vertex of the function, as a quadratic function can either have a minimum or maximum point. The vertex of the function can be found using the formula x = -b/(2a), where a, b, and c are the coefficients of the quadratic equation (in this case, a = 1, b = -2, and c = 2).
Using the formula, x = -(-2) / (2 * 1) = 1. Plugging this value back into the function, we can find the corresponding y-coordinate:
f(1) = 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1
So, the vertex of the function is (1, 1). Since the coefficient of the x^2 term is positive, the parabola opens upward, and the vertex represents the minimum point of the function.
This means that the range (R) of the function is all y-values greater than or equal to the y-coordinate of the vertex, which is 1.
Therefore, the correct answer is A. D = all real numbers and R = {y | y ≥ 1}.
To determine the domain and range of a function, we need to understand what these terms mean.
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In simpler terms, it represents all the values that x can take in the function.
The range of a function is the set of all possible output values (y-values) that the function can produce. In simpler terms, it represents all the values that y can take in the function.
To find the domain of the function f(x) = x^2 - 2x + 2, we need to consider any potential restrictions in the function. In this case, there are no restrictions to consider. Since this is a quadratic function, it is defined for all real numbers. Therefore, the domain is all real numbers.
Moving on to the range, we have to examine the behavior of the function. A quadratic function with no restrictions, like the one given, opens upward. Since there is no restriction on the x-values, the parabola will have a minimum or vertex at the bottom.
To find the range, we look at the y-coordinate of the vertex. In this case, the quadratic function is in the form f(x) = ax^2 + bx + c, where a = 1, b = -2, and c = 2. The x-coordinate of the vertex is found using the formula x = -b / (2a). Plugging in the values, we get x = -(-2) / (2 * 1) = 1.
To find the y-coordinate of the vertex, substitute this x-value back into the function: f(1) = 1^2 - 2(1) + 2 = 1 - 2 + 2 = 1. Therefore, the y-coordinate of the vertex (and the minimum y-value) is 1.
From this, we can conclude that the range of the function f(x) = x^2 - 2x + 2 is all values greater than or equal to 1. So, option A is correct: D = all real numbers, and R = {y | y ≥ 1}.