Suppose that we have an equation $y=ax^2+bx+c$ whose graph is a parabola with vertex $(3,2)$, vertical axis of symmetry, and contains the point $(1,0)$. What is $(a, b, c)$?
+128474 If the point (1,0) is on the graph, then by symmetry with the veryex, the point (5,0) is also on the graph
Then x =1 and x = 5 are roots
So......we have this form
y = a (x - 1) ( x - 5)
And since the point (3, 2) is on the graph, we can solve for "a" thusly
2 = a( 3 - 1) (3 - 5) simplify
2 = a (2) (-2)
2= - 4a divide both sides by -4
2/- 4 = a = -1/2
So we have that
y = (-1/2)(x - 1) (x - 5)
y = (-1/2) (x^2 - 6x + 5)
y = (-1/2)x^2 + 3x - 5/2
(a, b, c) = [ (-1/2) , 3, ( -5/2) ]
Here's the graph : https://www.desmos.com/calculator/vbzsmxqvnq
