Why is the y intercept for this equation positive?Rational function ( x + 1 ) ( 3 x + 2 ) ( x + 1 ) 2 I have found that the two asymptotes are: y = 3 and x = − 1 and there are no turning points for this function.To find the x intercept: 0 = ( x + 1 ) ( 3 x + 2 ) ( x + 1 ) 2 Hence 3 x 2 − x − 2 = 0 x 1 = 1 x 2 = − 2 3 Simularly to find the y intercept I did the following: ( 0 + 1 ) ( 3 ( 0 ) + 2 ) ( 0 + 1 ) 2 y = − 2However plotting this into a graphing software shows that the y intercept is y = 2 and x intercept is only x = − 2 3 and not x = 1.Why are the intercepts incorrect?

Question

Why is the y intercept for this equation positive?Rational function                    (        x        +        1        )                    (        3        x        +        2        )                            (        x        +        1        )                2      I have found that the two asymptotes are:   y  =  3 and   x  =  −  1 and there are no turning points for this function.To find the   x intercept:  0  =                    (        x        +        1        )                    (        3        x        +        2        )                            (        x        +        1        )                2      Hence  3      x    2    −  x  −  2  =  0      x    1    =  1               x    2    =  −      2    3  Simularly to find the   y intercept I did the following:                    (        0        +        1        )                    (        3        (        0        )        +        2        )                            (        0        +        1        )                2        y  =  −  2However plotting this into a graphing software shows that the   y intercept is   y  =  2 and   x intercept is only   x  =  −      2    3   and not   x  =  1.Why are the intercepts incorrect?

Jenna Farmer · Accepted Answer

(1)                                          (            x            +            1            )            (            3            x            +            2            )                                (            x            +            1                          )              2                                      =                              3            x            +            2                                x            +            1                              and that therefore the only zero of your function is located at   −      2    3  . On the other hand, if   x  =  0, then   (  1  ) is equal to   2.Also, using your approach, there is no need to solve a quadratic equation using the quadratic formula, since   (  3  x  +  2  )  (  x  +  1  )  =  0 if and only if   x  =  −      2    3   or   x  =  −  1. And   −  1 does not belong to the domain of your function.

Why is the y intercept for this equation positive? Rational function (

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