anonymous
  • anonymous
Finding the domain and range
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I need help finding the domian and range for my graph for y=3(x-2)^2+1
Z4K4R1Y4
  • Z4K4R1Y4
this should help: http://www.purplemath.com/modules/fcns2.htm
anonymous
  • anonymous
@Nnesha hey

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anonymous
  • anonymous
@Nnesha could you help?
anonymous
  • anonymous
Hey Nnesha you helped me last time. I was wondering if you were familiar with this?
Nnesha
  • Nnesha
hey blue alright so are u allowed to graph it ?
anonymous
  • anonymous
yeah i have to graph it ( i did already) now i have to find the domain and range
Nnesha
  • Nnesha
domain of parabola is always going to be negative inf to positive inf|dw:1433021151849:dw|
Nnesha
  • Nnesha
|dw:1433021252047:dw|we don't know how longer its going to be so that's why domain is \[(- ∞, ∞)\]
Nnesha
  • Nnesha
now range is y value so |dw:1433021354845:dw| and range start from the vertex(y point) to positive ∞
Nnesha
  • Nnesha
the red line shows starting point of range
anonymous
  • anonymous
okay so the range is (positive inf,1)
anonymous
  • anonymous
no its
Nnesha
  • Nnesha
starting point come first (start, end) so you don't know where its going to end so positive infinity
anonymous
  • anonymous
[1,positive inf)
Nnesha
  • Nnesha
yep right
anonymous
  • anonymous
okay great I understand it now. one more question
Nnesha
  • Nnesha
okay
Nnesha
  • Nnesha
|dw:1433021812495:dw|
anonymous
  • anonymous
okay yeah so i list those points?
Nnesha
  • Nnesha
parabola graph isn't crossing x-axis so no x-intercept is there any point where line cross y -axis ?
anonymous
  • anonymous
No
Nnesha
  • 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
  • anonymous
okay one sec
anonymous
  • anonymous
my vertex for this problem is (2,1) my graph looks different from yours
Nnesha
  • Nnesha
oh yeah it's that's just an example (weird drawing ik ik ;D)
Nnesha
  • Nnesha
|dw:1433022493243:dw| find this blue point where line cross y-axis
anonymous
  • anonymous
LOL okay
anonymous
  • anonymous
|dw:1433022652761:dw| my graph literally shows like this because the points that i plotted were (1,4) and (3,4)
Nnesha
  • Nnesha
plotted point why did you plot points ?
Nnesha
  • 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
  • anonymous
okay i see. im doing this by hand, so my graph only goes up to 5 lol. oops
anonymous
  • anonymous
Okay now lastly
Nnesha
  • Nnesha
ohhh okay so then you have to plug in 0 for x bec when line cross y-axis x value is 0
Nnesha
  • Nnesha
^^^ to find y-intercept b hand
anonymous
  • 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
  • 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
  • Nnesha
s*
Nnesha
  • Nnesha
in other words f(x)=x^2 is a parent function
anonymous
  • 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
  • Nnesha
lol glad to hear that :-) yes i'm not QH and btw you didn't ask QH question :-)
anonymous
  • anonymous
Well thank you so much for taking the time out to help me, I really appreciate.
Nnesha
  • Nnesha
my pleasure :-) thanks!

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