To determine the domain and range of the function [tex]\( y = (x + 3)^2 - 5 \)[/tex], let's analyze it step by step.
### Domain
The domain of a function refers to all the possible input values (x-values) for which the function is defined. Since the given function is a quadratic function in the form of [tex]\( y = (x + 3)^2 - 5 \)[/tex], there are no restrictions on the values that [tex]\( x \)[/tex] can take. Any real number can be substituted for [tex]\( x \)[/tex] without causing any undefined behavior like division by zero or taking the square root of a negative number. Therefore, the domain of this function is all real numbers:
[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.
1. Consider the vertex form of the quadratic function [tex]\( y = a(x - h)^2 + k \)[/tex]. For [tex]\( y = (x + 3)^2 - 5 \)[/tex], we can identify [tex]\( a = 1 \)[/tex], [tex]\( h = -3 \)[/tex], and [tex]\( k = -5 \)[/tex].
2. The vertex of this parabola is at [tex]\( (-3, -5) \)[/tex]. This is the minimum point of the function because the parabola opens upwards (since the coefficient of [tex]\( (x + 3)^2 \)[/tex] is positive).
At the vertex ([tex]\( x = -3 \)[/tex]):
[tex]\[ y = (-3 + 3)^2 - 5 = 0 - 5 = -5 \][/tex]
3. For values of [tex]\( x \)[/tex] different from [tex]\( -3 \)[/tex], the expression [tex]\( (x + 3)^2 \)[/tex] is always non-negative (i.e., [tex]\( (x + 3)^2 \geq 0 \)[/tex]), and thus [tex]\( (x + 3)^2 - 5 \geq -5 \)[/tex].
4. Therefore, the range of [tex]\( y = (x + 3)^2 - 5 \)[/tex] is all values of [tex]\( y \)[/tex] starting from [tex]\(-5\)[/tex] up to [tex]\( \infty \)[/tex].
So, the range is:
[tex]\[ \text{Range: } [-5, \infty) \][/tex]
Given these steps and descriptions, the correct answer is:
### C. Domain: [tex]\( (-\infty, \infty) \)[/tex]
### Range: [tex]\( [-5, \infty) \)[/tex]
