Sure, let's go through the steps to find the function [tex]\( h(x) \)[/tex].
### Step 1: Identify the Given Function
We are given the function:
[tex]\[ h(x) = \frac{1}{5x - 80} \][/tex]
### Step 2: Understand the Structure of the Function
Here, the function [tex]\( h(x) \)[/tex] is defined as the reciprocal of a linear expression in [tex]\( x \)[/tex]. Specifically, the denominator is [tex]\( 5x - 80 \)[/tex]. This means as long as [tex]\( 5x - 80 \neq 0 \)[/tex], [tex]\( h(x) \)[/tex] is defined.
### Step 3: Find the Domain of the Function
To find the domain, we must determine where the function is undefined. The function [tex]\( h(x) \)[/tex] will be undefined when the denominator is zero because division by zero is undefined. So, we set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 5x - 80 = 0 \][/tex]
[tex]\[ 5x = 80 \][/tex]
[tex]\[ x = 16 \][/tex]
Hence, the function [tex]\( h(x) \)[/tex] is defined for all [tex]\( x \)[/tex] except [tex]\( x = 16 \)[/tex].
So, the domain of [tex]\( h(x) \)[/tex] is:
[tex]\[ x \in (-\infty, 16) \cup (16, \infty) \][/tex]
### Step 4: Analyze the Behavior Around [tex]\( x = 16 \)[/tex]
Around [tex]\( x = 16 \)[/tex], the function tends toward either [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex] depending on the direction approached:
- As [tex]\( x \)[/tex] approaches 16 from the left (i.e., [tex]\( x \to 16^{-} \)[/tex]), [tex]\( 5x - 80 \)[/tex] approaches zero from the negative side, so [tex]\( h(x) \)[/tex] approaches [tex]\( -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches 16 from the right (i.e., [tex]\( x \to 16^{+} \)[/tex]), [tex]\( 5x - 80 \)[/tex] approaches zero from the positive side, and [tex]\( h(x) \)[/tex] approaches [tex]\( +\infty \)[/tex].
### Step 5: Conclusion
After analyzing the function, we can summarize the function as:
[tex]\[ h(x) = \frac{1}{5x - 80} \][/tex]
It is defined for all real numbers except [tex]\( x = 16 \)[/tex]. The domain of [tex]\( h(x) \)[/tex] is [tex]\( x \in (-\infty, 16) \cup (16, \infty) \)[/tex], and it has a vertical asymptote at [tex]\( x = 16 \)[/tex].
Thus, we've determined the nature and properties of the function [tex]\( h(x) \)[/tex].
