SOLUTION
The zeros pf the function are
[tex]-\frac{4}{3}\text{ and 1}[/tex]
But since the curve hits the x -axis and bounces, it an extra zero of 1.
That means the function will be in the form
[tex]y=a(x+\frac{4}{3})(x-1)(x-1)[/tex]
Note that the (x - 1) is reapeted.
Hence, it is a cubic function.
Since the y-intercept is 6, we set y to 6 and x to 0, and find a in the equation above, we have
[tex]\begin{gathered} y=a(x+\frac{4}{3})(x-1)(x-1) \\ 6=a(x+\frac{4}{3})(x-1)(x-1) \\ 6=a(0+\frac{4}{3})(0-1)(0-1) \\ 6=a(\frac{4}{3})(-1)(-1) \\ 6=\frac{4}{3}a \\ a=\frac{6}{\frac{4}{3}} \\ a=\frac{9}{2} \end{gathered}[/tex]
Hence, the answer is
[tex]y=\frac{9}{2}(x+\frac{4}{3})(x-1)(x-1)[/tex]
