Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
To determine the end behavior of the function [tex]\( h(x) = 2(x-3)^2 \)[/tex], we analyze what happens to the function values as [tex]\( x \)[/tex] approaches negative infinity and positive infinity.
Let's start with the end behavior as [tex]\( x \)[/tex] approaches negative infinity:
1. Consider the expression inside the function: [tex]\( (x-3)^2 \)[/tex].
2. For [tex]\( x \)[/tex] approaching negative infinity, [tex]\( x \)[/tex] becomes a very large negative number.
3. Subtracting 3 from a very large negative [tex]\( x \)[/tex] still results in a very large negative number.
4. Squaring this negative number [tex]\( (x-3)^2 \)[/tex] results in a very large positive number because the square of any real number is non-negative.
5. Multiplying the squared term by 2, which is a positive constant, maintains the positive nature and scales it.
6. Hence, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] will become a very large positive number.
Now, let's consider the end behavior as [tex]\( x \)[/tex] approaches positive infinity:
1. Again consider the expression inside the function: [tex]\( (x-3)^2 \)[/tex].
2. For [tex]\( x \)[/tex] approaching positive infinity, [tex]\( x \)[/tex] becomes a very large positive number.
3. Subtracting 3 from a very large positive [tex]\( x \)[/tex] still results in a very large positive number.
4. Squaring this positive number [tex]\( (x-3)^2 \)[/tex] results in a very large positive number.
5. Multiplying the squared term by 2, which is a positive constant, maintains the positive nature and scales it.
6. Hence, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] will also become a very large positive number.
Therefore, the correct statements are:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
So the end behavior of the function [tex]\( h(x) = 2(x-3)^2 \)[/tex] is:
[tex]\[ \begin{aligned} & \text{As } x \text{ approaches negative infinity, } h(x) \text{ approaches } \quad \infty, \\ & \text{As } x \text{ approaches positive infinity, } h(x) \text{ approaches } \quad \infty. \end{aligned} \][/tex]
Let's start with the end behavior as [tex]\( x \)[/tex] approaches negative infinity:
1. Consider the expression inside the function: [tex]\( (x-3)^2 \)[/tex].
2. For [tex]\( x \)[/tex] approaching negative infinity, [tex]\( x \)[/tex] becomes a very large negative number.
3. Subtracting 3 from a very large negative [tex]\( x \)[/tex] still results in a very large negative number.
4. Squaring this negative number [tex]\( (x-3)^2 \)[/tex] results in a very large positive number because the square of any real number is non-negative.
5. Multiplying the squared term by 2, which is a positive constant, maintains the positive nature and scales it.
6. Hence, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] will become a very large positive number.
Now, let's consider the end behavior as [tex]\( x \)[/tex] approaches positive infinity:
1. Again consider the expression inside the function: [tex]\( (x-3)^2 \)[/tex].
2. For [tex]\( x \)[/tex] approaching positive infinity, [tex]\( x \)[/tex] becomes a very large positive number.
3. Subtracting 3 from a very large positive [tex]\( x \)[/tex] still results in a very large positive number.
4. Squaring this positive number [tex]\( (x-3)^2 \)[/tex] results in a very large positive number.
5. Multiplying the squared term by 2, which is a positive constant, maintains the positive nature and scales it.
6. Hence, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] will also become a very large positive number.
Therefore, the correct statements are:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( h(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
So the end behavior of the function [tex]\( h(x) = 2(x-3)^2 \)[/tex] is:
[tex]\[ \begin{aligned} & \text{As } x \text{ approaches negative infinity, } h(x) \text{ approaches } \quad \infty, \\ & \text{As } x \text{ approaches positive infinity, } h(x) \text{ approaches } \quad \infty. \end{aligned} \][/tex]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
