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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- Jul 8th 2007, 02:43 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. - Jul 8th 2007, 02:54 AMjanvdl
For a)

$\displaystyle f(x) = a(x - \alpha)(x - \beta) $

$\displaystyle f(x) = a(x - 8)(x + 2) $

$\displaystyle f(x) = a(x^2 - 6x - 16) $

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For b)

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

$\displaystyle f(x) = (x^2 - 6x - 16) $

$\displaystyle \frac{dy}{dx} = 2x - 6 = 0 $

$\displaystyle x = 3 $

Set $\displaystyle x = 3 $ into $\displaystyle f(x) = (x^2 - 6x - 16) $

Then $\displaystyle f(3) = -25 $

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

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For c)

Now this one really isn't hard.

Look at this:

Would you be able to solve this?

$\displaystyle k^2 - 6k - 16 = 0 $

Of course you would!

But look at what i did:

I simply treated $\displaystyle (x + 3) $ as $\displaystyle k $

So let's solve this one:

$\displaystyle k^2 - 6k - 16 = 0 $

$\displaystyle (k - 8)(k + 2) = 0 $

$\displaystyle k = 8 \ and \ k = -2 $

But remember that $\displaystyle k $ is actually equal to $\displaystyle (x + 3) $

So $\displaystyle (x + 3) = 8 $ and $\displaystyle (x + 3) = -2 $

$\displaystyle x = 5 $ or $\displaystyle x = -5 $ - Jul 8th 2007, 07:36 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.

$\displaystyle y = x^2 - 6x - 16$

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

For vertex, $\displaystyle x = \frac {-b}{2a} = \frac {6}{2} = 3$

when $\displaystyle x = 3$

$\displaystyle y = 3^2 - 6(3) - 16 = -25$

So the coordinates for the lowest point is $\displaystyle (3,-25)$