Answer :
To determine the domain and range of the quadratic function [tex]\( y = 3x^2 - 6x + 5 \)[/tex], we need to analyze its behavior and properties.
### Domain
The domain of a function is the set of all possible input values ([tex]\(x\)[/tex]-values) for which the function is defined. Since [tex]\( y = 3x^2 - 6x + 5 \)[/tex] is a polynomial function, and polynomials are defined for all real numbers:
[tex]\[ \text{Domain} = (-\infty, \infty) \quad \text{or alternatively stated,} \quad \text{D: all real numbers} \][/tex]
### Range
The range of the function is the set of all possible output values ([tex]\(y\)[/tex]-values). To find the range of this quadratic function, we will identify the vertex of the parabola, as this will give us the minimum or maximum value of the function depending on the direction the parabola opens.
Given the quadratic function in standard form [tex]\( y = ax^2 + bx + c \)[/tex]:
[tex]\[ y = 3x^2 - 6x + 5 \][/tex]
The coefficient [tex]\( a \)[/tex] (which is 3) is positive, indicating that the parabola opens upwards. When a parabola opens upwards, the vertex represents the minimum point of the function.
The [tex]\( x \)[/tex]-coordinate of the vertex is calculated using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting [tex]\( a = 3 \)[/tex] and [tex]\( b = -6 \)[/tex]:
[tex]\[ x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1 \][/tex]
Next, we substitute [tex]\( x = 1 \)[/tex] back into the function to find the corresponding [tex]\( y \)[/tex]-value (the minimum value of [tex]\( y \)[/tex]):
[tex]\[ y = 3(1)^2 - 6(1) + 5 = 3 - 6 + 5 = 2 \][/tex]
Therefore, the vertex of the parabola is at the point [tex]\( (1, 2) \)[/tex]. Since the parabola opens upwards, the minimum value of [tex]\( y \)[/tex] is 2. This means that [tex]\( y \)[/tex] can take any value greater than or equal to 2. Thus, the range of the function is:
[tex]\[ \text{Range} = [2, \infty) \quad \text{or alternatively stated as} \quad R: (y \geq 2) \][/tex]
### Conclusion
From the above analysis:
- The domain of the function is all real numbers.
- The range of the function is [tex]\( y \geq 2 \)[/tex].
Thus, the correct answer is:
a. D: all real numbers, R: (y \geq 2)
### Domain
The domain of a function is the set of all possible input values ([tex]\(x\)[/tex]-values) for which the function is defined. Since [tex]\( y = 3x^2 - 6x + 5 \)[/tex] is a polynomial function, and polynomials are defined for all real numbers:
[tex]\[ \text{Domain} = (-\infty, \infty) \quad \text{or alternatively stated,} \quad \text{D: all real numbers} \][/tex]
### Range
The range of the function is the set of all possible output values ([tex]\(y\)[/tex]-values). To find the range of this quadratic function, we will identify the vertex of the parabola, as this will give us the minimum or maximum value of the function depending on the direction the parabola opens.
Given the quadratic function in standard form [tex]\( y = ax^2 + bx + c \)[/tex]:
[tex]\[ y = 3x^2 - 6x + 5 \][/tex]
The coefficient [tex]\( a \)[/tex] (which is 3) is positive, indicating that the parabola opens upwards. When a parabola opens upwards, the vertex represents the minimum point of the function.
The [tex]\( x \)[/tex]-coordinate of the vertex is calculated using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting [tex]\( a = 3 \)[/tex] and [tex]\( b = -6 \)[/tex]:
[tex]\[ x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1 \][/tex]
Next, we substitute [tex]\( x = 1 \)[/tex] back into the function to find the corresponding [tex]\( y \)[/tex]-value (the minimum value of [tex]\( y \)[/tex]):
[tex]\[ y = 3(1)^2 - 6(1) + 5 = 3 - 6 + 5 = 2 \][/tex]
Therefore, the vertex of the parabola is at the point [tex]\( (1, 2) \)[/tex]. Since the parabola opens upwards, the minimum value of [tex]\( y \)[/tex] is 2. This means that [tex]\( y \)[/tex] can take any value greater than or equal to 2. Thus, the range of the function is:
[tex]\[ \text{Range} = [2, \infty) \quad \text{or alternatively stated as} \quad R: (y \geq 2) \][/tex]
### Conclusion
From the above analysis:
- The domain of the function is all real numbers.
- The range of the function is [tex]\( y \geq 2 \)[/tex].
Thus, the correct answer is:
a. D: all real numbers, R: (y \geq 2)