At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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 hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
