how do i find a if my equation is y=a(x-.5)^2+2
Do you have a known value of x and y?
No. I just have my vertex for h,k which is (0.5,2)
do you know any other points on this parabola?
yes (0.5,2) & (7,2)
your vertex is certainly a point on the parabola, no?
Yes
so plug in the coordinates of a known point and solve for \(a\)
Thats where im confused. The equation in general form is y=a(x-h)^2+k my h,k can either be (7,2) or (0.5,2) so i have now y=a(x-7)^2+2 ?
y=a(x-.5)^2+2 2=a(7-0.5)^2+2 ... plug in (7,2) 2=a(6.5)^2+2 ... ... ... a = ???
A=0.04?
oh wait, are you sure the other point (that's not the vertex) is (7,2) ?
The two points on the paraobla i have is (.5, 2) & (7,2)
then none of those two are the vertex
So what is the h,k then ?
the vertex would be the point (h,k) where h = (0.5+7)/2 = 7.5/2 = 3.75 but we can't determine the value of k unless we get more info
What else information do i need
another point at least
(9,2)(2.5,2.08)(5.68,5.1)
are you able to post a picture of this at all?
How do i post it?
click attach file
under the text box
Where?
see where it says "draw"?
im on my phone i dont see it
oh, hmm I'm not sure if it's the same on mobile devices
try scrolling down?
I can send the link
ok
Its not letting me , but the picture is on google search real world parabolas And its the sixth picture
all I see is the mcdonalds picture and a bunch of roller coasters...hmm I'm doing something wrong
No its the tower picture after the red roller coaster
so this page http://mathforum.org/mathimages/index.php/Parabola ???
No go to google image search & type real world parabolas
this is what I get http://cameronjtesblog.blogspot.com/2011/04/real-world-parabolas.html
so I think I'm on the wrong page (again...)
Go to google.com
ok
then what?
Then type in real world parabolas
ok did that
which # link am I looking for?
There's no link . Its a picture of a tower
yeah getting the same results, are you sure there's no way to send the link? the direct link
It wont copy right . Im trying
thanks
Do you see it
I get the same results that I did if I were to search it myself I meant the direct link to the actual picture...sorry for being a pain
I think your phone and my computer deliver different results or something, not sure
Its okay hold on.
ok
yeah that's what I've been getting...but that has no numbers on it
I'm guessing you're supposed to add in the points later?
yeah i use this geogebra thing & it gives me the points
ah send me that if you can as well, if not, then its ok
basically I was hoping to see this thing for myself to see where the vertex was
Im not close to a computer so i cant
alright then no worries
Okay
I can tell you right off the bat that the vertex will not be on the same horizontal level as any other point that's on the parabola
so there's no way that (0.5, 2) can be the vertex if another point is (7, 2)
So it would be the middle point
yeah at (3.75, k)
But how do i find k
and you also said that (9,2) is on the parabola right?
Yes
that's just not possible in any parabola, only 2 points will share the same y coordinate (so you can't have 3 points that have a y coordinate of 2)
so this is part of the reason for my confusion and why I wanted to see a pic of the whole thing
I have to make a quadratic formula based of the parabola in the picture & the geogebra gaves me those points
hmm there's probably a typo somewhere then if we had the 3 points (0,5, 2) (7, 2) and (9,2) the you would either have a linear equation or a cubic (but not a quadratic) going through those 3 points
so hopefully you see the issue here
no the points are (9,2)(2.5,2.0)(5.68,5.1)
I'm assuming those points are labeled A, B, and C in geogebra right?
Yeah
ok what you can do is with geogebra you can type "fitpoly[A,B,C,2]" without quotes into the input bar at the bottom what this will do is interpolate a polynomial (ie draw a polynomial through all 3 points) and it will force the polynomial to be of degree 2, so it will be a parabola
this saves you a lot of time from doing it by hand (which involves ugly systems of equations and matrices), so that's fortunate you're able to use geogebra
Okay. And then from there what do i do
what equation do you get?
when you type that command into geogebra
I didnt do i just wanted to know so that when i do i wont get stuck
well once you have the equation, you can determine the value of 'a' by looking at the leading coefficient
you can also determine the vertex by typing in "max[f,-20,20]" without quotes into the input bar
Okay, & then.from there im asked to find the x & y-intercepts
the y-intercept would be the last number in the function you get from after doing the "fitpoly" command the x-intercepts can be found by typing "root[f]" without quotes into the input bar
Okay thanks
you're welcome
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