Question 1190538: Find the quadratic function y=a(x−h)2 whose graph passes through the given points.
(4,−2) and (2,0)
Found 3 solutions by ikleyn, Alan3354, greenestamps:
Answer by ikleyn(52779)   (Show Source):
You can put this solution on YOUR website! .
Find the quadratic function y=a(x−h)2 whose graph passes through the given points.
(4,−2) and (2,0)
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Let's consider the point (2,0) first.
The fact that it belongs to the given form parabola MEANS that
0 = a*(2-h)^2
which, with a=/=0, IMPLIES h= 2.
So, half of the problem is just solved, and we know that h= 2.
THEREFORE, we can re-write the quadratic function in the form
y = a*(x-2)^2.
Now consider the second point, (4,-2).
The fact that it belongs to the given form parabola MEANS that
-2 = a*(4-2)^2, or -2 = a*2^2, or -2 = 4a
which implies a=  = -0.5.
So, the quadratic function is y = -0.5*(x-2)^2, and h= 2, a= -0.5. ANSWER
Solved.
Answer by Alan3354(69443)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The equation is  , which is equivalent to  ; the vertical shift is 0, so the vertex is on the x-axis.
One of the given points is (2,0), which is on the x-axis, so (2,0) is the vertex.
The other given point is below the x-axis, so the graph opens downward; the leading coefficient "a" is negative.
To determine the leading coefficient, we can use formal mathematics as the other tutor did; but we can find it informally, as follows:
The second point given is 2 units to the right of the vertex. If the leading coefficient were -1, the y coordinate of that second point would be -(2^2) = -4; but it is only -2. That means the leading coefficient is -1/2 instead of -1.
So a=-1/2 and h=2.
ANSWER: 
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