ANSWERS
Function:
[tex]h(x)=\frac{x+1}{(x-1)(x+3)}[/tex]Key features:
• Asymptotes: x = 1, x = -3, (vertical), and ,y = 0, (horizontal).
,• Intercepts: (0, -1/3), (-1, 0)
,• Symmetry: none
,• Example of end behavior: as x → 1⁺, y → infinity,, while, as x → 1⁻, y → -infinity
EXPLANATION
The vertical asymptotes of a rational function are given by the zeros of the denominator. For example, the function,
[tex]h(x)=\frac{x+1}{(x-1)(x+3)}[/tex]Has two vertical asymptotes: one at x = 1, and one at x = -3.
The horizontal asymptote is determined by the degrees of the numerator (n) and denominator (m):
• If n < m then the horizontal asymptote is the x-axis
,• If n = m then the horizontal asymptote is y = a/b, where a and b are the leading coefficients of the numerator and denominator, respectively.
,• If n > m then there is no horizontal asymptote
In the given example, the degree of the numerator is 1, while the degree of the denominator is 2. Thus, function h(x) has a horizontal asymptote that is the x-axis.
Now, we have to find the key features for this function:
• y-intercept:, occurs when x = 0
[tex]h(0)=\frac{0+1}{(0-1)(0+3)}=\frac{1}{-3}=-\frac{1}{3}[/tex]Hence, the y-intercept is (0, -1/3)
• x-intercepts:, the x-intercepts are the x-intercepts of the numerator: ,(-1, 0),.
,• The ,asymptotes, are the ones mentioned above: ,x = 1, x = -3, (vertical), and ,y = 0, (horizontal).
,• The ,symmetry, is determined as follows:
[tex]\begin{gathered} Even\text{ }functions:f(-x)=f(x) \\ Odd\text{ }functions:f(-x)=-f(x) \end{gathered}[/tex]If we replace x with -x in function h(x), we will find that the result is neither h(x) nor -h(x) and, therefore this function is neither even nor odd.
Finally, an example of the end behavior of this function around the asymptote x = 1 is that as x → 1⁺, y → infinity, while as x → 1⁻¹, y → -infinity.
If we take values of x greater than x = 1 - so we will approach 1 from the right, we will get values of the function that show an increasing behavior. Let's find h(x) for x = 4, x = 3, and x = 2,
[tex]\begin{gathered} h(4)=\frac{4+1}{(4-1)(4+3)}=\frac{5}{3\cdot7}=\frac{5}{21} \\ \\ h(3)=\frac{3+1}{(3-1)(3+3)}=\frac{4}{2\times6}=\frac{4}{12}=\frac{1}{3} \\ \\ h(2)=\frac{2+1}{(2-1)(2+3)}=\frac{3}{1\times5}=\frac{3}{5} \end{gathered}[/tex]And we will find the opposite behavior for values that are less than 1 - but greater than -1 because there the function has a zero and its behavior could change,
[tex]\begin{gathered} h(0)=\frac{0+1}{(0-1)(0+3)}=\frac{1}{-1\times3}=-\frac{1}{3} \\ \\ h(\frac{1}{2})=\frac{\frac{1}{2}+1}{(\frac{1}{2}-1)(\frac{1}{2}+3)}=\frac{\frac{3}{2}}{-\frac{1}{2}\times\frac{7}{2}}=\frac{\frac{3}{2}}{-\frac{7}{4}}=-\frac{3\cdot4}{2\cdot7}=-\frac{6}{7} \end{gathered}[/tex]The graph of the function and its key features is,
