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To determine the end behavior of the function [tex]\( h(x) = 2(x-3)^2 \)[/tex], let's analyze what happens as [tex]\( x \)[/tex] approaches both positive and negative infinity.
1. Expression Analysis:
- The function [tex]\( h(x) = 2(x-3)^2 \)[/tex] is a quadratic function in the form of [tex]\( a(x-b)^2 + c \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 0 \)[/tex].
2. Behavior as [tex]\( x \)[/tex] Approaches Positive Infinity:
- As [tex]\( x \)[/tex] becomes very large (towards positive infinity), the expression [tex]\( x - 3 \)[/tex] will also become very large.
- Squaring a large positive or negative number results in a very large positive number.
- Multiplying this large positive number by 2 will keep it positive and large.
- Hence, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( 2(x-3)^2 \)[/tex] will also approach positive infinity.
3. Behavior as [tex]\( x \)[/tex] Approaches Negative Infinity:
- As [tex]\( x \)[/tex] becomes very large in the negative direction (towards negative infinity), the expression [tex]\( x - 3 \)[/tex] will also become very large in the negative direction.
- Squaring a large negative number also results in a very large positive number.
- Multiplying this large positive number by 2 will still keep it positive and large.
- Hence, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( 2(x-3)^2 \)[/tex] will again approach positive infinity.
Therefore, we can conclude:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches positive infinity.
Thus, the correct answer is:
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches [tex]\(\boxed{\infty}\)[/tex].
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches [tex]\(\boxed{\infty}\)[/tex].
1. Expression Analysis:
- The function [tex]\( h(x) = 2(x-3)^2 \)[/tex] is a quadratic function in the form of [tex]\( a(x-b)^2 + c \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 0 \)[/tex].
2. Behavior as [tex]\( x \)[/tex] Approaches Positive Infinity:
- As [tex]\( x \)[/tex] becomes very large (towards positive infinity), the expression [tex]\( x - 3 \)[/tex] will also become very large.
- Squaring a large positive or negative number results in a very large positive number.
- Multiplying this large positive number by 2 will keep it positive and large.
- Hence, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( 2(x-3)^2 \)[/tex] will also approach positive infinity.
3. Behavior as [tex]\( x \)[/tex] Approaches Negative Infinity:
- As [tex]\( x \)[/tex] becomes very large in the negative direction (towards negative infinity), the expression [tex]\( x - 3 \)[/tex] will also become very large in the negative direction.
- Squaring a large negative number also results in a very large positive number.
- Multiplying this large positive number by 2 will still keep it positive and large.
- Hence, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( 2(x-3)^2 \)[/tex] will again approach positive infinity.
Therefore, we can conclude:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches positive infinity.
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches positive infinity.
Thus, the correct answer is:
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches [tex]\(\boxed{\infty}\)[/tex].
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches [tex]\(\boxed{\infty}\)[/tex].
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