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Consider the end behavior of this function, and then complete the statements.

[tex]\[ h(x) = -\frac{1}{5}|x-3| + 4 \][/tex]

As \( x \) approaches negative infinity, \( h(x) \) approaches \_\_\_\_\_.

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

Sagot :

GinnyAnsweridn
To analyze the end behavior of the function \( h(x) = -\frac{1}{5}|x-3| + 4 \), let's consider what happens as \( x \) approaches negative infinity and positive infinity.

### Step-by-Step Analysis:
1. Structure of the Function:
- The function consists of two parts: \(-\frac{1}{5}|x-3|\) and the constant term \( +4 \).

2. Behavior of \(|x-3|\) as \(x\) approaches negative infinity:
- As \( x \) approaches negative infinity, the expression \(|x-3|\) implies the absolute value of a large negative number minus 3, which becomes a large positive number because the absolute value of a large negative number is large.
- Hence, \(|x-3|\) tends towards positive infinity.

3. Behavior of \(-\frac{1}{5}|x-3|\) as \(x\) approaches negative infinity:
- Since \(|x-3|\) tends towards positive infinity, multiplying this large number by \(-\frac{1}{5}\) will result in a term that tends towards negative infinity.
- \(-\frac{1}{5}|x-3|\) thus tends towards negative infinity.

4. Combining with the Constant \( +4 \):
- Adding 4 to a term that tends towards negative infinity still gives a value that tends towards negative infinity.

Therefore, as \( x \) approaches negative infinity, \( h(x) \) approaches negative infinity.

5. Behavior as \( x \) approaches positive infinity:
- The steps are similar to when \( x \) approaches negative infinity because \(|x-3|\) will again tend towards positive infinity.
- Multiplying this large positive number by \(-\frac{1}{5}\) will also result in negative infinity.
- Adding the constant 4 does not change the end behavior significantly.

Therefore, as \( x \) approaches positive infinity, \( h(x) \) also approaches negative infinity.

### Conclusion:
- As \( x \) approaches negative infinity, \( h(x) \) approaches \( -\infty \).
- As \( x \) approaches positive infinity, \( h(x) \) also approaches \( -\infty \).

So, the completions for the statements are:
- As \( x \) approaches negative infinity, \( h(x) \) approaches \( -\infty \).
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( h(x) \)[/tex] approaches [tex]\( -\infty \)[/tex].
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