Finding the domain and range

- anonymous

Finding the domain and range

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- schrodinger

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- anonymous

I need help finding the domian and range for my graph for y=3(x-2)^2+1

- Z4K4R1Y4

this should help:
http://www.purplemath.com/modules/fcns2.htm

- anonymous

@Nnesha hey

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## More answers

- anonymous

@Nnesha could you help?

- anonymous

Hey Nnesha you helped me last time. I was wondering if you were familiar with this?

- Nnesha

hey blue
alright so are u allowed to graph it ?

- anonymous

yeah i have to graph it ( i did already) now i have to find the domain and range

- Nnesha

domain of parabola is always going to be negative inf to positive inf|dw:1433021151849:dw|

- Nnesha

|dw:1433021252047:dw|we don't know how longer its going to be so that's why
domain is \[(- ∞, ∞)\]

- Nnesha

now range is y value so |dw:1433021354845:dw|
and range start from the vertex(y point) to positive ∞

- Nnesha

the red line shows starting point of range

- anonymous

okay so the range is (positive inf,1)

- anonymous

no its

- Nnesha

starting point come first
(start, end) so you don't know where its going to end so positive infinity

- anonymous

[1,positive inf)

- Nnesha

yep right

- anonymous

okay great I understand it now. one more question

- Nnesha

okay

- Nnesha

|dw:1433021812495:dw|

- anonymous

okay yeah so i list those points?

- Nnesha

parabola graph isn't crossing x-axis so no x-intercept
is there any point where line cross y -axis ?

- anonymous

No

- Nnesha

mhm yes there is a point where line cross y-axis
if you r using desmos caluclator zoomout a little bit so you can see

- anonymous

okay one sec

- anonymous

my vertex for this problem is (2,1) my graph looks different from yours

- Nnesha

oh yeah it's that's just an example (weird drawing ik ik ;D)

- Nnesha

|dw:1433022493243:dw|
find this blue point where line cross y-axis

- anonymous

LOL okay

- anonymous

|dw:1433022652761:dw|
my graph literally shows like this because the points that i plotted were (1,4) and (3,4)

- Nnesha

plotted point why did you plot points ?

- Nnesha

http://prntscr.com/7b9y0p
this is your graph do you see the point where line cross y-axis between 10 and 15
you need that point for y-intercept
zoomout a little bit to see :-)

- anonymous

okay i see. im doing this by hand, so my graph only goes up to 5 lol. oops

- anonymous

Okay now lastly

- Nnesha

ohhh okay
so then you have to plug in 0 for x
bec when line cross y-axis x value is 0

- Nnesha

^^^ to find y-intercept b hand

- anonymous

the main question is for me to compare f(x)= a(x-h)^2 +k to f(x) x^2.
i explained that they are basically the same its just that the h and k show us how left/right/up/down the curve has shifted. does that make sense?

- Nnesha

y = x^2 is a basic form of quadratic function
it shows the shape of parabola
yes you are right
the difference
f(x)= a(x-h)^2 +k is a vertex form equation
where h represent horizontal shift
and k is for vertical shift
where a is tell how how wider or Skinner the graph is going to be

- Nnesha

s*

- Nnesha

in other words f(x)=x^2 is a parent function

- anonymous

Okay great YAY! I feel smart
Thank you so much for helping me better understand, I wish I could repay you. Your not a qualified helper right? :-(

- Nnesha

lol glad to hear that :-)
yes i'm not QH and btw you didn't ask QH question :-)

- anonymous

Well thank you so much for taking the time out to help me, I really appreciate.

- Nnesha

my pleasure :-)
thanks!

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