# Hi there :) Here's the question (I'm studying for my Pre-Calculus Final) Sketch the graph of y = !0.5(x-2)^2 - 4! note: the exclaimation marks (!) are representative of absolute value...

Hi there :)

Here's the question (I'm studying for my Pre-Calculus Final)

Sketch the graph of y = !0.5(x-2)^2 - 4!

note: the exclaimation marks (!) are representative of absolute value lines, because I don't know how to put in the absolute value lines here :P

Thanks in advance!! :D

### 1 Answer | Add Yours

Sketch the graph of `y=|.5(x-2)^2-4|` :

If `f(x)=|a|` then `f(x)=a,a>0` and `f(x)=-a,a<0`

So the graph will be the graph of the parabola `y=.5(x-2)^2-4` except where this parabola is below the x-axis -- for this interval(s) the graph will be the reflection over the x-axis.

Now the graph of `y=.5(x-2)^2-4` is a parabola opening up, so we need only worry about any x values where y<0. If there are any such x values they must occur between the two zeros. So we find the zeros:

`1/2(x-2)^2-4=0`

`1/2(x^2-4x+4)-4=0`

`1/2x^2-2x+2-4=0`

`1/2x^2-2x=2`

`x^2-4x=4`

`x^2-4x+4=8`

`(x-2)^2=8`

`x-2=+-sqrt(8)=+-2sqrt(2)`

`x=2+-2sqrt(2)`

So for `2-2sqrt(2)<x<2+2sqrt(2)` the graph will be a reflection about the x-axis. (The y-values will be `-[.5(x-2)^2-4]` )

The graph of `y=.5(x-2)^2-4` is a parabola opening up, vertex at (2,-4) -- this parabola has been vertically compressed compared to `y=x^2` )

The graph:

The green dotted line is `y=.5(x-2)^2-4` (no absolute value.)