SOLUTION: 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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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(52780)   (Show Source): You can put this solution on YOUR website!
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Find the quadratic function y=a(x−h)2 whose graph passes through the given points.
​(4​,−2​) and ​(2​,0)
~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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): You can put this solution on YOUR website!
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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