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What is the end behavior of the function [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 :

To determine the end behavior of the function [tex]\( h(x) = 2(x-3)^2 \)[/tex], we need to analyze how the function behaves as [tex]\( x \)[/tex] approaches both negative infinity and positive infinity.

Step-by-Step Analysis:

1. Understand the function's form: The given function is
[tex]\[ h(x) = 2(x-3)^2. \][/tex]
This 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. Characteristics of quadratic functions:
- The quadratic function [tex]\( y = a(x - h)^2 + k \)[/tex] is a parabola.
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards.

3. Given [tex]\( a = 2 \)[/tex] which is positive: Thus, the parabola [tex]\( h(x) \)[/tex] opens upwards.

4. End behavior as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
- Analyze [tex]\( h(x) \)[/tex] as [tex]\( x \to -\infty \)[/tex]:
[tex]\[ (x - 3)^2 \text{ grows very large because } (x - 3) \text{ becomes a large negative number squared}. \][/tex]
- Since [tex]\((x - 3)^2\)[/tex] is always non-negative and grows without bound as [tex]\( x \)[/tex] moves towards [tex]\(-\infty\)[/tex]:
[tex]\[ 2(x - 3)^2 \text{ also grows without bound (multiplied by 2)}. \][/tex]
- Therefore, [tex]\( h(x) \)[/tex] approaches [tex]\( +\infty \)[/tex].

5. End behavior as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex]:
- Analyze [tex]\( h(x) \)[/tex] as [tex]\( x \to +\infty \)[/tex]:
[tex]\[ (x - 3)^2 \text{ grows very large because } (x - 3) \text{ becomes a large positive number squared}. \][/tex]
- Again, [tex]\((x - 3)^2\)[/tex] is always non-negative and grows without bound as [tex]\( x \)[/tex] moves towards [tex]\( +\infty \)[/tex]:
[tex]\[ 2(x - 3)^2 \text{ also grows without bound (multiplied by 2)}. \][/tex]
- Therefore, [tex]\( h(x) \)[/tex] approaches [tex]\( +\infty \)[/tex].

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

- 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].

Correct answers for the drop-down menus:

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].
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