Answer :
Let's analyze the function \( f(x) = 2(x + 6)^2 + 5 \) to determine its domain and range.
### Domain:
The domain of a function refers to all the possible input values (x-values) for which the function is defined. In this case, \( f(x) = 2(x + 6)^2 + 5 \) is a quadratic function.
For any real number \( x \), you can calculate \( f(x) \), because you can always square a real number and then multiply and add constants to it. There are no restrictions like division by zero or taking the square root of a negative number.
Therefore, the domain of \( f(x) \) is all real numbers, which we can write in interval notation as:
[tex]$ \text{Domain: } (-\infty, \infty) $[/tex]
### Range:
The range of a function refers to all the possible output values (y-values) that the function can produce.
To find the range, we need to consider the nature of the quadratic function \( f(x) = 2(x + 6)^2 + 5 \). This function is written in vertex form \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. Here, the vertex form is \( f(x) = 2(x + 6)^2 + 5 \), with \( a = 2 \), \( h = -6 \), and \( k = 5 \).
Since \( a = 2 \) is positive, the parabola opens upwards. The minimum value of \( f(x) \) occurs at the vertex \( x = -6 \) and is \( f(-6) = 2(-6 + 6)^2 + 5 = 5 \).
Because the parabola opens upwards and the minimum value of \( f(x) \) is 5, the function can take any value greater than or equal to 5. Therefore, the range of \( f(x) \) in interval notation is:
[tex]$ \text{Range: } [5, \infty) $[/tex]
So, summarizing our results:
[tex]$ \boxed{\text{Domain: } (-\infty, \infty) \quad \text{Range: } [5, \infty)} $[/tex]
### Domain:
The domain of a function refers to all the possible input values (x-values) for which the function is defined. In this case, \( f(x) = 2(x + 6)^2 + 5 \) is a quadratic function.
For any real number \( x \), you can calculate \( f(x) \), because you can always square a real number and then multiply and add constants to it. There are no restrictions like division by zero or taking the square root of a negative number.
Therefore, the domain of \( f(x) \) is all real numbers, which we can write in interval notation as:
[tex]$ \text{Domain: } (-\infty, \infty) $[/tex]
### Range:
The range of a function refers to all the possible output values (y-values) that the function can produce.
To find the range, we need to consider the nature of the quadratic function \( f(x) = 2(x + 6)^2 + 5 \). This function is written in vertex form \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. Here, the vertex form is \( f(x) = 2(x + 6)^2 + 5 \), with \( a = 2 \), \( h = -6 \), and \( k = 5 \).
Since \( a = 2 \) is positive, the parabola opens upwards. The minimum value of \( f(x) \) occurs at the vertex \( x = -6 \) and is \( f(-6) = 2(-6 + 6)^2 + 5 = 5 \).
Because the parabola opens upwards and the minimum value of \( f(x) \) is 5, the function can take any value greater than or equal to 5. Therefore, the range of \( f(x) \) in interval notation is:
[tex]$ \text{Range: } [5, \infty) $[/tex]
So, summarizing our results:
[tex]$ \boxed{\text{Domain: } (-\infty, \infty) \quad \text{Range: } [5, \infty)} $[/tex]
