Let the polynomial be y = ax^2 + bx + c
The derivative or slope at the minimum is zero: 2ax + b = 0
We have 55 = a (4^2) + b (4) + c
25 = a (3^2) + b (3) + c
and 2a(3) + b = 0 or b = -6a
55 = 16a - 24a + c = -6a + c
25 = 9a -18a +c = - 9a + c
30 = 3a
a = 10
b = -60
c = 55 + 6a = 115
The parabola is y = 10x^2 - 60x + 115
Check using, for instance
http://www.wolframalpha.com/input/?i=evaluate+y+%3D+10x^2+-+60x+%2B+115+at+x%3D3
and
http://www.wolframalpha.com/input/?i=minimize++10x^2+-+60x+%2B+115
A vertex is the highest or lowest point in a parabola.
(y - 3) = a(x - 1)2 y = a(x - 1)2 + 3 4 = a(4 - 1)2 + 3 1 = 9a a = 1/9 y = 1/9 (x - 1)2 + 3
right
Above
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
-2
5
The coordinates will be at the point of the turn the parabola which is its vertex.
2
Go study
The vertex would be the point where both sides of the parabola meet.
you didn't put any equations, but the answer probably begins with y= (x-4)^2+1
The vertex -- the closest point on the parabola to the directrix.
A vertex is the highest or lowest point in a parabola.
A parabola is NOT a point, it is the whole curve.
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the right of the vertex.
The vertex, or maximum, or minimum.