# $f(x) = (x + 1)(x − 2)^2$ . a Sketch the curve $y = f(x)$, showing the coordinates of any points where the curve meets the coordinate axes.

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?

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

Hope this graph helps -

its of $f(x)=(x+1)(x-2)^2$

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

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