# Using translation and change of scale to sketch graphs of these quadratics

I borrowed the following problems from MIT open courseware problem sets. (This is the very first problem in the problem set.) I understand what completing the square is, but what does it mean when they say “use translation” and “change of scale”. I am unable to solve this as I didn't understand the question. It would be nice if I could get the solution.

By completing the square, use translation and change of scale to sketch:

a) $y = x^2 − 2x − 1$;

b) $y =3x^2 +6x +2$.

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Note that $y=(x-1)^2-2$. Now draw $y=x^2$, pull it over to the right by $1$, then down by $2$. – André Nicolas Sep 5 '11 at 19:46
How did get y = (x-1)^2 - 2,please show the steps – alok Sep 5 '11 at 20:04
I completed the square. You had indicated you knew about that. We have $x^2-2x -1=(x-1)^2-1-1=(x-1)^2-2$. For the other problem, you will look at $3[x^2+2x+2/3]=3[(x+1)^2-1/3]=3(x+1)^2-1$. Graph $y=x^2$, move it left by $1$, scale in $y$-direction by factor of $3$, then move the result down by $1$. – André Nicolas Sep 5 '11 at 20:26
@alok: For a) look at this picture with the parabola $y=x^2$ (black) and the parabola $y=x^2-2x-1=(x-1)^2-2$ (blue) translated as André Nicolas indicated. – Américo Tavares Sep 5 '11 at 20:45

## 1 Answer

For a)

Here is a picture with the parabola $y=x^2$ (black) and the parabola $y=x^2−2x−1=(x−1)^2−2$ (blue) translated as André Nicolas indicated in a comment above.

(Blue axes: $X=x-1$ is the dashed horizontal axis and $Y=y+2$ is the dashed vertical axis. See below)

I present my own explanation.

If you had the parabola $y=x^{2}$ you would be able to sketch its graph. The idea is to get a parabola in the same form, but translated with respect to it. How?

By completing the square you write $y=x^2-2x-1$ as

$$\color{blue}{y}=x^2-2x+1-2=\color{blue}{\left( x-1\right) ^{2}-2}.$$

Now, if you translate the $x,y$-axes, by making the change of variables $x=X+1$ and $y=Y-2$, you get

$$\color{blue}{Y=X^{2}},$$

which is a parabola centered at $(X,Y)=(0,0)$. Note that the point $(X,Y)=(0,0)$ is located at $\color{blue}{(x,y)=(1,-2)}$. In the figure the dash lines are the $X$- and $Y$-axes, while the solid ones are the $x$- and $y$-axes.

In this case you have no need to change the scale, because the coefficient of $x^2$ is $1$.

Added: For b)

$$\begin{eqnarray*} \color{blue}y &=&3x^{2}+6x+2=3\left( x^{2}+2x+\frac{2}{3}\right) \\ &=&3\left( x^{2}+2x+1-\frac{1}{3}\right) =3\left( \left( x+1\right) ^{2}- \frac{1}{3}\right) \\ &=&\color{blue}{3\left( x+1\right) ^{2}-1}. \end{eqnarray*}$$

If you put $x=X-1,y=Y-1$, you get $$\color{blue}{Y=3X^{2}}.$$ The point $(X,Y)=(0,0)$ is now $\color{blue}{(x,y)=(-1,-1)}$ and the factor $3$ is the coefficient of $x^2$.

Plots of $y=3x^2$ (black) and $y=3x^2+6x+2=3(x+1)^2-1$ (blue)

(Blue axes: $X=x+1$ is the dashed horizontal axis and $Y=y+1$ is the dashed vertical axis.)

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Special Thanks to all of you and a very special thanks to Srivatsan and Americo for the solution and also my sincere appologies for not getting back on time because of Internet issues in my locality. – alok Oct 30 '11 at 9:15