sketching mod functions

h(x) = 2(X+3)^2 -8
sketch the curve with the equation y= h(-|x|)

I am struggling.
The book says, for h(|x|) to sketch the graph of h(x) for x>= 0. That's a quadratic curve starting at y=10,

Then it says to reflect it in the y-axis. So then there'd be a V ish shape. Then if I did -ve of that graph it would look identical to what I have.

The book's answer is the original curve + its reflection from y= -8 t0 10.


How come?

h(x) = 2(X+3)^2 -8
sketch the curve with the equation y= h(-|x|)

I am struggling.
The book says, for h(|x|) to sketch the graph of h(x) for x>= 0. That's a quadratic curve starting at y=10,

Then it says to reflect it in the y-axis. So then there'd be a V ish shape. Then if I did -ve of that graph it would look identical to what I have.

The book's answer is the original curve + its reflection from y= -8 t0 10.


How come?

If would help to see your curve/the book answer/the original question, but mr desmos he says
https://www.desmos.com/calculator/aii7ctf4ho
which matches?

Just to add a small tip here. When needing to sketch a modulus graph whose shape is not immediately obvious, it can help to build the graph up via a series of thumbnail sketches, in this case:

mod(x)
-mod(x)
-mod(x) + 3
(-mod(x) + 3)^2
2(-mod(x) + 3)^2
2(-mod(x) + 3)^2 - 8

As you can see this is what I try to do but the mod x bit troubles me in this particular question

Here is the question and answer

Maybe try something similar but starting with mod(x) rather than applying the mod function later.

So the key thing is to sketch h(x) for negative x (the -|x| part) then reflect in the y axis (as its |x|) which is what they seem to have done.

This is from the aqa text book from which the question came. What am I mis interpreting in my working

As per the previous post you want to sketch h(x) when x is negative, then reflect in the y axis. A quick check of the value of the function for x=-1 and 2 say
* h(-|-1|) = h(-|1|) = h(-1). So when x is negative h(-|x|) = h(x) so sketch the quadratic h(x) for x<0
* h(-|2|) = h(-2). So when x is positive h(-|x|) = h(-x). So reflect the quadratic in the y axis for x>0

OK So when its -|X| you disregard the graph to the right of the y-axis and then continue. Thanks

All good, but if stuck without a suitable rule of thumb in an exam you can always build towards the required graph using stages something like: