Parabola Equation Help :x
So I'm in dire need for someone to help me come up with an equation with the following information given :l
I'm having a quarterly coming up soon and the teacher hinted on having 2 of these types of questions on the test. I do not have a clue as to how to even start, and I was hoping someone may clear it up for me nicely!
The information given is:
Axis: Parallel to y-axis
Passes through the points:
(0,3) , (3,4) , (4,11)
Directions are to find the equation of the specified parabola.
If someone could show me the steps as well that would be greatly appreciated.
November 12, 2010
11 Comments • Newest first
I explained everything I could. Look through my posts and do everything I said. There should be no uncertainty on how to do this problem.
[quote=radkai]I go to Stuyvesant High School. (aka the top high school of New York City, google/wikipedia can prove this)[/quote]
I know what it is. I go to Horace Mann lol.
[quote=hyj1106]What kinda school do you go to? We don't have resource websites lol[/quote]
I go to Stuyvesant High School. (aka the top high school of New York City, google/wikipedia can prove this)
[quote=Billionz]If the textbook doesn't explain everything, never forget to use the resource websites that the school has provided for you.[/quote]
What kinda school do you go to? We don't have resource websites lol
If the textbook doesn't explain everything, never forget to use the resource websites that the school has provided for you.
[quote=PandorasBox]About the (0,3) being the lowest point? o.o[/quote]
Of course.
[quote=PandorasBox]It's pretty simple really.
(0,3) is the vertex of the parabola because you can see those two other points are on one side of the axis and continue to grow while (0,3) is the lowest point.
Then graph (3,4) and (4,11). Then procede to flip them over the y-axis and draw the parabola.
As for the equation of a vertex equation of a parabola is: a(x-h)^2+k. Where -h = x and k=y of the vertex. I don't really know how you would find "a" or how to put it into standard form but I think that's a good start for you.[/quote]
Incorrect assumption!
@hyj1106 dude.... a=5/3 b=-14/3 Put those numbers in the standard form of a quadratic equation. 5x^2/3-14x/3+3=y
[quote=radkai]You don't know how to do systems? o.o 1st equation you get 1=9a+3b. 2nd you get 2=4a+b. Multiply the 2nd equation by -3 to get -6=-12a-3b. Combine the two equations. -5=-3a. a=5/3 Plug 5/3 back into any equation. 2=20/3 +b. b=-14/3[/quote]
OKay I got that much done, I did know how to do them but I wasn't exactly sure.
Uhm, so how would I proceed after this?
[quote=hyj1106]The given information is
Axis: Parallel to y-axis
Passes through the points: (0,3), (3,4), and (4,11)
Exactly verbatim. :x
@Above: what do I do next? I"m seriously clueless[/quote]
You don't know how to do systems? o.o 1st equation you get 1=9a+3b. 2nd you get 2=4a+b. Multiply the 2nd equation by -3 to get -6=-12a-3b. Combine the two equations. -5=-3a. a=5/3 Plug 5/3 back into any equation. 2=20/3 +b. b=-14/3
The given information is
Axis: Parallel to y-axis
Passes through the points: (0,3), (3,4), and (4,11)
Exactly verbatim. :x
@Above: what do I do next? I"m seriously clueless
y=ax^2+bx+c c=3 because it passes through (0,3) now plug in the other points. 4=9a+3b+3 (3,4) 11=16a+4b+3 (4,11) you got system of equations