help, will fan and medal!! What is the a value of the following function?
???????
idk, i dont get what its asking
the function is x^2 but this question is incomplete....
so it would be 2?
I think so.....
that isn't simply a x^2 function.
@phi not given in question..... i already said the question is incomplete @stacy.reed08
based on the graph, I'd say the function itself is \[\frac{ 1 }{ 3 }x ^{2}\]
ok, after re-reading they are asking about the "a" value I assume they mean, if we write the equation in the form y= a(x-h)^2 + k they want the "a"
hmmm.....
(h,k) is the vertex (the lowest point) and for your parabola, the vertex is (0,0) that means h=0 and k=0 and the equation is y = a(x-0)^2 + 0 and that simplifies to y= a x^2 to find a, put in an (x,y) pair that is on the parabola. For example, try using (3,3)
so, its not 2..... hmmmm that's sure....
@phi , nice catch! I completely missed the "a" in the problem. It makes much more sense now :)
I would definitely use the point (3,3) since it's the only point that is clearly at an intersection of 2 whole numbers. Just plug it into the equation and solve for a. \[y=ax ^{2}\]\[3=a \times3^{2}\]\[3=a \times9\] finish it out and you're done!
im still confused..i been sitting here trying to figure this out..
Use the vertex form for finding the equation of a parabola. @phi already provided this... (h,k) is the vertex, which for the graph you provided is (0,0). Pick a point that is easy to identify like (3,3) and plug it into the vertex form equation. \[y=a(x-h)^{2}+k\]
Plug in what you know... Given the point (3,3), y=3 and x=3. Given the vertex (0,0), h=0 and k=0 you end up with the equation below...just solve for "a" \[3=a(3-0)^{2}+0\]
3=9a /9 3/9=a 1/3=a
got it!
i wish i could give you both medals..
@phi definitely deserves it for catching the part we were all missing in the original problem!
I will help you to give medal to one of them. You pick one to do, I do the rest :)
I gave it to @phi, but @stacy.reed08 i have more confussing ones that i could use some help one.
@Penguin7 , I'll try my best! Just tag me in the problem.
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