Question

How do you graph `y= 1( x-3)^2 -3`?

Answer 1

Showing how to determine the critical points. Needed for graph sketching. For a more precise plot will need to build a table of additional points.

Explanation:

`color(blue)("Important preamble")`

This is the vertex form equation of a quadratic in that

`y=ax^2+bx+c` is the same thing as `y=a(x+b/(2a))^2+c+k`

Where `k` is a correction for the introduced error consequential to 'forcing' `y=ax^2+bx+c` into this format.

Note that `a(b/(2a))^2+k=0`

The advantage of this vertex type equation is that with a little 'tweaking' you can virtually read of the coordinates for the vertex. Hence the name of 'vertex' form equation.

`x_("vertex")=(-1)xxb/(2a)`

`y_("vertex")=c+k`
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Given that `y=1(x-3)^2-3` we have the following:

`color(blue)("Determining the critical points - the general shape")`

`color(blue)("As "a->1" is positive the general shape of the graph is "uu)`

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`color(blue)("Determining the critical points - the vertex")`

Given the equation:

`y=a(x+b/(2a))^2+c+k" "->" "y=1(xcolor(magenta)(-3))^2color(green)(-3)`

`x_("vertex")=(-1)xx(color(magenta)(-3))=+3`

`y_("vertex")=color(green)(-3)`

`=>color(blue)(" Verertex "->(x,y)=(+3,-3))`
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`color(blue)("Determining the critical points - the y-intercept")`

Expanding the backets:

`y=(x-3)^2-3" "->" "y=x^2-6x+9-3`

`" "->" "color(green)(y=x^2-6xcolor(red)(+6))`

`" "y_("intercept")=color(red)(+6)`
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`color(blue)("Determining the critical points - the x-intercept")`

The x-axis is at `y=0`. So the graph crosses the x-axis at `y=0`

`y=(x-3)^2-3" "->" "0=(x-3)^2-3`

Add 3 to both sides

`3=(x-3)^2`

Take the square root of both sides

`+-sqrt(3)=x-3`

Add 3 to both sides

`+-sqrt(3)+3=x`

`x~~4.732 and x~~1.268` to 3 decimal places
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If you need any more points you will need to build a table.
Tony B