Questions: Determine the domain and range of f(x)=6(x+3)^2+1. Enter the solution in interval notation. - Domain: - Range:

Solution Steps

To determine the domain and range of the function \( f(x) = 6(x+3)^2 + 1 \):

1. Domain: The domain of a quadratic function is all real numbers because there are no restrictions on the values that \( x \) can take.
2. Range: The range of the function can be found by analyzing the vertex of the parabola. Since the coefficient of the squared term is positive, the parabola opens upwards. The vertex form of the quadratic function \( f(x) = a(x-h)^2 + k \) indicates that the minimum value of the function is \( k \) when \( a > 0 \).

The function \( f(x) = 6(x + 3)^2 + 1 \) is a quadratic function. Quadratic functions are defined for all real numbers. Therefore, the domain is:

\[ \text{Domain: } (-\infty, \infty) \]

To find the range, we first identify the vertex of the parabola. The vertex occurs at \( x = -3 \). Substituting this value into the function gives:

\[ f(-3) = 6(-3 + 3)^2 + 1 = 6(0)^2 + 1 = 1 \]

Since the coefficient of the squared term is positive, the parabola opens upwards, indicating that the minimum value of the function is at the vertex. Thus, the range is:

\[ \text{Range: } [1, \infty) \]

Final Answer