Question

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

Answer 1

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 `(x_1+1,y_1-1)`

Explanation:

`color(magenta)("Preamble")`

As the coefficient of `x^2` is positive `(+1x^2)` then the graph is of form `uu`. Thus the vertex is a minimum.

`color(red)("If")` the coefficient had been negative then the graph would have been in the form `nn`. Thus the vertex would have been a maximum.
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`color(magenta)("Answering the question")`

What we are transforming is the basis of `y=x^2` where `x_("vertex")=0`
Let the vertex of `y=x^2->(x_1,y_1)=(0,0)`

Note this is the same as: `y=x^2+0x+0`
Note that the y-intercept is at `x=0`
So for this case the y-intercept is `y=(0)^2+0x+0=0`

`color(blue)("Transformation left or right - Shift left or right")`

Let the vertex of `y=x^2-2x ->(x_2,y_2)`

By including the `color(red)(-2)color(green)(x)` the new `x_("vertex")` of `color(green)(y=x^2color(red)(-2)x)` is `(-1/2)xxcolor(red)(-2)=+1 =x_2`

So the transformation for `x` is `x_2-x_1" "=" "1-0=+1`

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`color(blue)("Transformation up or down - shift up or down")`

`y_(vertex)" for "y=x^2=0=y_1`

The new `y_("vertex") =y_2` at `x_2=1`

So by substitution for `x` `y_2=(x_2)^2-2(x_2)" "=" "(1)^2-2(1)=-1`

Thus transformation for y is `y_2-y_1" "=" "-1-0" "=" "-1`
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Consequently the transformation of any point is `(x_1+1,y_1-1)`