Let's consider the quadratic function given by the equation [tex]\( f(x) = 2(x-4)^2 - 2 \)[/tex]. The goal is to determine the domain and the range of this function.
### Domain:
The domain of a function is the set of all possible input values (x-values) that the function can accept. For the given function [tex]\( f(x) = 2(x-4)^2 - 2 \)[/tex], which is a quadratic function, there are no restrictions on the values of [tex]\( x \)[/tex]. Quadratic functions are defined for all real numbers. However, in this specific problem, there is a constraint provided: [tex]\( x \geq -2 \)[/tex]. Therefore, the domain of the function is:
[tex]\[ x \geq -2 \][/tex]
or in interval notation:
[tex]\[ [-2, \infty) \][/tex]
### Range:
The range of a function is the set of all possible output values (y-values) that the function can produce. Since the given function is of the form [tex]\( f(x) = 2(x-4)^2 - 2 \)[/tex], we can analyze it further:
- The basic form of the quadratic function is [tex]\( (x-4)^2 \)[/tex], which attains its minimum value of [tex]\( 0 \)[/tex] when [tex]\( x = 4 \)[/tex].
- Multiplying by 2 scales the function vertically, but does not affect the minimum value location. Therefore, the function [tex]\( 2(x-4)^2 \)[/tex] still attains its minimum value [tex]\( 0 \)[/tex] when [tex]\( x = 4 \)[/tex].
- Subtracting 2 translates the function downward by 2 units.
Thus, the minimum value of the function [tex]\( f(x) = 2(x-4)^2 - 2 \)[/tex] is [tex]\( -2 \)[/tex], and it occurs at [tex]\( x = 4 \)[/tex].
Since the parabola opens upwards (because the coefficient of the squared term [tex]\( 2 \)[/tex] is positive), the function can take any value greater than or equal to [tex]\( -2 \)[/tex]. Therefore, the range of the function is:
[tex]\[ y \geq -2 \][/tex]
or in interval notation:
[tex]\[ [-2, \infty) \][/tex]
### Conclusion:
The domain and range of the quadratic function [tex]\( f(x) = 2(x-4)^2 - 2 \)[/tex] are as follows:
- Domain: [tex]\([ -2, \infty )\)[/tex]
- Range: [tex]\([ -2, \infty )\)[/tex]
