Question 571736: A square centered at the origin has its vertices on the x and y axes. The graph of the function
f(x) = ax2 - 4, a > 0
passes through three of the square's vertices. Find a.
I tried to draw a graph of it, but didn't really help, since I don't know the exact measures of the square... But being that its vertices are on the x and y axes I figured that it is like, diagonal, but other than that, I am completely lost!
Please help, I would definitely appreciate it :)
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Let the four vertices of the square (starting at the negative  axis and going clockwise) be ) , ) , ) , ) .
Since the second degree term variable in the function is we know that the axis of symmetry of the parabola graph is a vertical line. Furthermore, since we are given that  , we know that the parabola opens upward, and the vertex is a minimum local extrema. That fixes the three points of the square that are in common with the graph of the function as  ,  , and  with as the vertex of the parabola.
The standard form of a parabola is:
\ =\ ax^2\ +\ bx\ +\ c)
where 
So if the parabola passes through the point ) then
^2\ +\ b(x_i)\ +\ c)
In the case of your function at point ,
^2\ +\ (0)(0)\ -\ 4\ =\ -\beta)
from which we deduce
But since the given figure was a square, we know that
Subsituting the coordinates of which we now know to be ) into your given function:
^2\ +\ (0)(4)\ -\ 4\ =\ 0)
from which we get
John
My calculator said it, I believe it, that settles it
|
|
|
