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I tried putting $y=0$, then having $(x+1)$, $(x-2)$, $(x-2)$; where $x$ would equal $(0,..)$ respectively. Is that correct, and not sure what to sketch?

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  • 1
    $\begingroup$ 2.a bit before 'Sketch' needs editing out. $\endgroup$
    – user108815
    Nov 18, 2013 at 14:58
  • $\begingroup$ The abscissa(en.wikipedia.org/wiki/Abscissa) $=0$ for $y$ axis $\endgroup$ Nov 18, 2013 at 14:59
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    $\begingroup$ Is $f=(x+1)(x-2)^2$ the function you want to study? $\endgroup$
    – Avitus
    Nov 18, 2013 at 14:59
  • $\begingroup$ yes that is the function $\endgroup$
    – user108815
    Nov 18, 2013 at 15:03

1 Answer 1

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Hope this graph helps -

its of $f(x)=(x+1)(x-2)^2$ enter image description here

\\\\\\\\\\\\\EDIT\\\\\\\\\\\\\\\
Here you can substitute for $y=f(x)$ and then take different values of $x$ $$ \begin{array}{c|lcr} x & \text{y} \\ \hline -1 & (-1+1)(-1-2)^2=\color{blue}{0} \\ 1 & (1+1)(1-2)^2=\color{blue}{2} \\ 2 & (2+1)(2-1)^2=\color{blue}{0} \\ 3 & (3+1)(3-1)^2=\color{blue}{4} \\ \end{array} $$

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  • $\begingroup$ How did you get from the function to the graph? $\endgroup$
    – user108815
    Nov 18, 2013 at 16:51
  • $\begingroup$ @user108815 I have edited my answer to make it as requested $\endgroup$ Nov 19, 2013 at 9:53

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