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

Question

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?

The abscissa(http://en.wikipedia.org/wiki/Abscissa) $=0$ for $y$ axis — lab bhattacharjee, Nov 18 '13 at 14:59

Shivam Patel · Answer 1 · 2013-11-19T09:53:20.003

0

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 $x$ $$ \begin{array}{c|lcr} x & \text{y} \\ \hline -1 & (-1+1)(-1-2)^2=\color{blue}{0} \\ 1 & (1+1)(1-2)^2=\color{blue}{2} \\ 2 & (2+1)(2-1)^2=\color{blue}{0} \\ 3 & (3+1)(3-1)^2=\color{blue}{4} \\ \end{array} $$

edited Nov 19 '13 at 09:53

answered Nov 18 '13 at 15:14

Shivam Patel

4,012

How did you get from the function to the graph? – user108815 Nov 18 '13 at 16:51
@user108815 I have edited my answer to make it as requested – Shivam Patel Nov 19 '13 at 09:53

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

1 Answers1