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Re: Parabola [#permalink]
sandy wrote:
taureanamir wrote:

The equation of the parabola shown in the figure is y = 2x - x^2



The book says answer is B. B is greater. But how??



[img]
Attachment:
14233852_10153688722265563_1487696121_o.jpg
[/img]

Qty A
t

Qty B
2s-s^2




Hi taureanamir,

Quantity B can be rewritten as y as \(2s-s^2\) =\(y(s)\). Now from the figure t < y as point (s,t) is inside the parabola and x-axis.

Hence B is greater.

PS: Please share the source of the question.

Regards

Could you please make it more clear by elaborate?
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Re: Parabola [#permalink]
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AE wrote:
Could you please make it more clear by elaborate?


They say pictures are worth a 1000 words.

Attachment:
Inkedparabola_LI.jpg
Inkedparabola_LI.jpg [ 925.79 KiB | Viewed 14015 times ]



So you have been given an equation of parabola you can use the value of x coordinate to calculate the y coordinate. If you see the picture above it becomes clear why B is greater.
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Re: Parabola [#permalink]
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taureanamir wrote:
Thank you very much Sandy.

The question is from a local institute's book.

Could you please elaborate on the solution a little.


Hey there, I solved it a little differently. First found the vertex (1, 1): x coordinate of the vertex = -b/2a. Then plug in x and solve for y, which will give you the (x,y) coordinate of the vertex. In this case it is (1,1).


Since the point (s, t) is below the vertex (1,1), s<1 and t<1. From here I just picked a number that satisfied s<1 and t<1 and solved. It worked; there may be a better way to do this, but hope this helps a little.
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Re: The equation of the parabola shown in the figure above is [#permalink]
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There is a different way to solve this:

Let's assume there is a point on the parabola directly above the given point. The x coordinate will remain the same that is s. But the y coordinate will be something like t+c, where c is a positive number. So the coordinates of the point should be (s,t+c).

As this point is on the parabola it should satisfy the equation of the parabola so, plugin the values. It gives us t+c=2s-s2. Therefore if we subtract c from both sides we can see that t is c less than 2s-s2, as c is a positive number. So B is the correct choice.
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Re: The equation of the parabola shown in the figure above is [#permalink]
sandy wrote:
AE wrote:
Could you please make it more clear by elaborate?


They say pictures are worth a 1000 words.

Attachment:
Inkedparabola_LI.jpg



So you have been given an equation of parabola you can use the value of x coordinate to calculate the y coordinate. If you see the picture above it becomes clear why B is greater.



Thanks for this explanation. I now understand that the question does not even need you to perform any calculation.

s=x at the same point of the parabola but from the position it is less than y in the equation of the parabola.

Since s=x is constant and t is less than y, then B is the answer.
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Re: The equation of the parabola shown in the figure above is [#permalink]
Given : y = 2x−x^2 , we need to find the point (s,t) which is slightly less than the vertex of parabola.

There we are many ways, you can find the vertex but easiest will be using formula , let vertex (p , q)

for p = -b/2a (formula) , Given : y = 2x−x^2 which is same as y = ax^2 + bx + c
a= -1 and b = 2 so, p = -2/-2 = 1 , put value of p (i.e x) in parabola equation y = 2x−x^2 = 2 -1 = 1

so (p, q) = (1,1) as (s,t) slightly less than vertex take (s,t) = (0.7, 0.7)

t = 0.7 s= 2(0.7) - (0.7)^2 = 0.91

thus B
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