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What is the end behavior of the function [tex]h[/tex]?
[tex]\[ h(x)=2(x-3)^2 \][/tex]

As [tex]x[/tex] approaches negative infinity, [tex]h(x)[/tex] approaches [tex]$\square$[/tex].

As [tex]x[/tex] approaches positive infinity, [tex]h(x)[/tex] approaches [tex]$\square$[/tex].

Sagot :

GinnyAnswer
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]
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