Question

What are the important points needed to graph `y=3(x+3)^2 - 3`?

Answer 1

`y_("intercept")=24`
`x_("intercepts")=-4 and -2`
Vertex `->(x,y)=(-3,-3)`

Explanation:

This is the vertex form of the standardised`" " y=ax^2+bx+c`
Multiplying out the brackets the we have `" "y=3x^2+18x+24`

As the `3x^2` is positive the graph is of general shape `uu`

`color(blue)("Determine the vertex using the questions equation format")`

Given:`" "y=3(xcolor(red)(+3))^2color(green)(-3)`

`x_("vertex")=(-1)xxcolor(red)((+3)) = -3`
`y_("vertex")=color(green)(-3)`

Vertex`->(x,y)=(-3,-3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determine the x-intercepts")`

set `y=0`

`=>0=3(x+3)^2-3`

`3=3(x+3)^2`

`(x+3)^2=1`

`x+3=+-sqrt((1)`

`x=-3+-1`

`x=-4 and -2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine y-intercept")`

Set `x=0`

`y= 3(0+3)^2-3`

`y=27-3=24`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Foot note")`

Using the format `y=3x^2+18x+24`

`x_("vertex")=(-1/2)xx(18/3)=-3`