# How do you sketch the graph of y=x^2-2x and describe the transformation?

Jun 9, 2017

The graph of $y = {x}^{2}$ moves to the right by 1
The graph of $y = {x}^{2}$ moves down by 1

Thus the transformation of any point is $\left({x}_{1} + 1 , {y}_{1} - 1\right)$

#### Explanation:

$\textcolor{m a \ge n t a}{\text{Preamble}}$

As the coefficient of ${x}^{2}$ is positive $\left(+ 1 {x}^{2}\right)$ then the graph is of form $\cup$. Thus the vertex is a minimum.

$\textcolor{red}{\text{If}}$ the coefficient had been negative then the graph would have been in the form $\cap$. Thus the vertex would have been a maximum.
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$\textcolor{m a \ge n t a}{\text{Answering the question}}$

What we are transforming is the basis of $y = {x}^{2}$ where ${x}_{\text{vertex}} = 0$
Let the vertex of $y = {x}^{2} \to \left({x}_{1} , {y}_{1}\right) = \left(0 , 0\right)$

Note this is the same as: $y = {x}^{2} + 0 x + 0$
Note that the y-intercept is at $x = 0$
So for this case the y-intercept is $y = {\left(0\right)}^{2} + 0 x + 0 = 0$

$\textcolor{b l u e}{\text{Transformation left or right - Shift left or right}}$

Let the vertex of $y = {x}^{2} - 2 x \to \left({x}_{2} , {y}_{2}\right)$

By including the $\textcolor{red}{- 2} \textcolor{g r e e n}{x}$ the new ${x}_{\text{vertex}}$ of $\textcolor{g r e e n}{y = {x}^{2} \textcolor{red}{- 2} x}$ is $\left(- \frac{1}{2}\right) \times \textcolor{red}{- 2} = + 1 = {x}_{2}$

So the transformation for $x$ is ${x}_{2} - {x}_{1} \text{ "=" } 1 - 0 = + 1$

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$\textcolor{b l u e}{\text{Transformation up or down - shift up or down}}$

${y}_{v e r t e x} \text{ for } y = {x}^{2} = 0 = {y}_{1}$

The new ${y}_{\text{vertex}} = {y}_{2}$ at ${x}_{2} = 1$

So by substitution for $x$ ${y}_{2} = {\left({x}_{2}\right)}^{2} - 2 \left({x}_{2}\right) \text{ "=" } {\left(1\right)}^{2} - 2 \left(1\right) = - 1$

Thus transformation for y is ${y}_{2} - {y}_{1} \text{ "=" "-1-0" "=" } - 1$
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Consequently the transformation of any point is $\left({x}_{1} + 1 , {y}_{1} - 1\right)$