Explanation:
comparing the two function you have:
[tex]\begin{gathered} 1)\text{ }f(x)\text{ = }\frac{2x(x\text{ - 2)(x + 1)}}{3(x\text{ + 2)(x - 3)(x + 1)}} \\ \\ In\text{ this function:} \\ x\text{ intercepts: x = 0 and x - 2 = 0} \\ x\text{ = 0 and x = 2} \\ \\ \text{The hole is the factor co}mmon\text{ to both numerator and denominator:} \\ (x\text{ + 1) is co}mmon\text{ to both numerator and denominator} \\ \\ \text{vertical asymptote: Equate denominator to zero:} \\ (x\text{ + 2) = 0 , (x - 3) = 0 , (x + 1) = 0 \lbrack(x+1) is a hole, so we won't include it} \\ \text{vertical asymptote: }x\text{ = -2, x = 3 } \\ \\ \text{horizontal asymptote = 2/3} \end{gathered}[/tex]2nd function:
[tex]\begin{gathered} 2)\text{ }f(x)\text{ = }\frac{2(x-2)^2\text{(x + 1)}}{3(x\text{ + 2)(x - 3)(x + 1)}} \\ \\ In\text{ this function:} \\ x\text{ intercept: (x - 2) twice} \\ x\text{ - 2= 0} \\ x\text{ intercept: x = 2} \\ \\ The\text{ hole is factor co}mmon\text{ to both numerator and denominator} \\ (x\text{ + 1) is co}mmon\text{ to both numerator and denominator} \\ x\text{ + 1 = 0} \\ x\text{ = -1 (is a hole)} \\ \\ \text{vertical asymptote: Equate denominator to zero} \end{gathered}[/tex][tex]\begin{gathered} \text{vertical asymptote: Equate denominator to zero} \\ (x\text{ + 2) = 0 , (x - 3) = 0 , (x + 1) = 0} \\ \text{ (x + 1) = 0 is a hole, so we won't include it} \\ \text{vertical asymptote: }x\text{ = -2, x = 3} \\ \\ \text{horizontal asymptote = 2/3} \end{gathered}[/tex]The difference between both functions is the x-intercept
The first function have two intercepts while the second function have only one x-intercept
From our question, the x intercept is supposed to be just 1
we were given the x-intercept as x = 2
Function has x -intercept as 2.
Function 2 is the correct representation of the rational function that gives those conditions in the question
[tex]f(x)\text{ = }\frac{2(x-2)^2\text{(x + 1)}}{3(x\text{ + 2)(x - 3)(x + 1)}}[/tex]