Question 352781: A parabola's vertex is at the point (10,0). If this parabola passes through the point (8,4), where must it cross the y-axis?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
First thing: Since the value of the function for the given point is greater than the value of the function at the vertex, the parabola must open upwards.
Second thing: By symmetry, since the parabola passes through (8,4) it must also pass through (12,4). (the  -value 8 is 2 smaller than the -value of the vertex, so the point with an -value 2 larger than the -value of the vertex must also have a function value of 4.)
Now we have three points and can determine the coefficients of the quadratic function.
We have a quadradic function of the form:
\ =\ ax^2\ +\ bx +\ c)
And three points on the graph of our function, (10,0), (8,4), and (12,4).
Considering the first point:
^2\ +\ b(10) +\ c\ =\ 0)
which is to say:
Likewise, from the coordinates of the other two points we can derive:
And
Since
has a  -intercept at ) , all you need to do is solve the system of equations to determine the value of 
John
My calculator said it, I believe it, that settles it
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