To determine the end behavior of the function [tex]\( f(x) = -\frac{1}{4} x^2 \)[/tex], we need to analyze how the function behaves as [tex]\( x \)[/tex] approaches positive infinity and negative infinity.
1. As [tex]\( x \rightarrow \infty \)[/tex]:
- Consider the term [tex]\(-\frac{1}{4} x^2\)[/tex].
- As [tex]\( x \)[/tex] becomes very large, [tex]\( x^2 \)[/tex] will become very large as well.
- Because [tex]\( x^2 \)[/tex] is positive for all [tex]\( x \)[/tex], multiplying by [tex]\(-\frac{1}{4}\)[/tex] means that the result will be negative.
- The larger [tex]\( x \)[/tex] gets, the more negative [tex]\(-\frac{1}{4} x^2\)[/tex] will become.
- Thus, as [tex]\( x \rightarrow \infty \)[/tex], [tex]\( f(x) \)[/tex] approaches [tex]\(-\infty \)[/tex].
2. As [tex]\( x \rightarrow -\infty \)[/tex]:
- Similarly, consider the term [tex]\(-\frac{1}{4} x^2\)[/tex].
- As [tex]\( x \)[/tex] becomes very large in the negative direction, [tex]\( x^2 \)[/tex] will again become very large.
- Even though [tex]\( x \)[/tex] is negative, [tex]\( x^2 \)[/tex] is still positive (since squaring a negative number gives a positive result).
- Multiplying this large positive [tex]\( x^2 \)[/tex] by [tex]\(-\frac{1}{4}\)[/tex] means that the result will be very negative.
- Therefore, the further [tex]\( x \)[/tex] gets from zero in the negative direction, the more negative [tex]\(-\frac{1}{4} x^2\)[/tex] will become.
- Hence, as [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \)[/tex] also approaches [tex]\(-\infty \)[/tex].
In summary:
- As [tex]\( x \rightarrow \infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
- As [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
The matching statements from the given options are:
1. As [tex]\( x \rightarrow \infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
2. As [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
