For (a); I tried this:

y = x^2 + qx + r

y = o^2 + 6 * 0 + 16

y = -16

y = x^2 -6x - 16

y = 8^2 - 6*8 - 16

y = 0

I am lost with (b) and not sure about (c)

Thanks.

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- July 8th 2007, 03:43 AMGAdamsEquation Graph
For (a); I tried this:

y = x^2 + qx + r

y = o^2 + 6 * 0 + 16

y = -16

y = x^2 -6x - 16

y = 8^2 - 6*8 - 16

y = 0

I am lost with (b) and not sure about (c)

Thanks. - July 8th 2007, 03:54 AMjanvdl
For a)

-----------

For b)

I think we can get the minimum point by using Calculus.

Set into

Then

So point P's coordinates is: (3 ; - 25)

-------------------

For c)

Now this one really isn't hard.

Look at this:

Would you be able to solve this?

Of course you would!

But look at what i did:

I simply treated as

So let's solve this one:

But remember that is actually equal to

So and

or - July 8th 2007, 08:36 AMJhevon
To (b)

I don't think GAdams has taken a calculus class as yet. We can use the vertex formula to get the lowest Point.

This is an upward opening parabola, so the vertex occurs at the minimum point.

For vertex,

when

So the coordinates for the lowest point is