Given:
• Vertical asymptote at : x = -1
,
• Double zero at: x = 2
,
• y-intercept at: (0, 2)
Let's create the equation for the rational function using the given properties.
Since the vertical asymptote is at x = -1, to find the deominator of the equation, equate the vertical asymptote to zero.
Add 1 to both sides:
[tex]\begin{gathered} x+1=-1+1 \\ x+1=0 \end{gathered}[/tex]
Therefore, the denominator of the function is ==> x + 1
Since it has a double zero at x = 2, we have the factors:
[tex]\Longrightarrow(x-2)(x-2)[/tex]
We now have the equation:
[tex]y=\frac{a(x-2)(x-2)}{x+1}[/tex]
Also, the y-intercept is at: (0, 2)
To find the value o a, substitute 2 for y and 0 for x then evaluate:
[tex]\begin{gathered} 2=\frac{a(0-2)(0-2)}{0+1} \\ \\ 2=\frac{a(-2)(-2)}{1} \\ \\ 4a=2 \\ \\ a=\frac{2}{4} \\ \\ a=\frac{1}{2} \end{gathered}[/tex]
Therefore, the rational function is:
[tex]y=\frac{\frac{1}{2}(x-2)(x-2)}{x+1}[/tex]
ANSWER:
[tex]y=\frac{\frac{1}{2}(x-2)(x-2)}{x+1}[/tex][tex]\begin{gathered} \text{Numerator: }\frac{1}{2}(x-2)(x-2) \\ \\ \\ \text{Denominator: (x + 1)} \end{gathered}[/tex]
