Parabola

**dgolverk** · November 29th 2006, 02:19 PM

Hey guys,
Can you help me answer this question?
*Find the equation, in vertex form, of the quadratic realtion with zeors at -2 and 8 and y-intercept of 8.

Thanks

**topsquark** · November 29th 2006, 03:08 PM

Originally Posted by dgolverk

Hey guys,
Can you help me answer this question?
*Find the equation, in vertex form, of the quadratic realtion with zeors at -2 and 8 and y-intercept of 8.

Thanks

You want the form:
$y = a(x - h)^2 + k$

You have zeros at (-2, 0) and (8, 0) and a y-intecept of (0, 8).

Plugging these points into the equation individually:
$0 = a(-2 - h)^2 + k$
$0 = a(8 - h)^2 + k$
$8 = a(0 - h)^2 + k$

Or:
$ah^2 + 4ah + 4a + k = 0$
$ah^2 - 16ah + 64a + k = 0$
$ah^2 + k = 8$

Probably the simplest thing to do here would be to take the last equation:
$ah^2 = 8 - k$ and sub it into the top two equations.

$(8 - k) + 4ah + 4a + k = 0$ ==> $4ah + 4a + 8 = 0$
$(8 - k) - 16ah + 64a + k = 0$ ==> $-16ah + 64a + 8 = 0$

These turn into:
$ah + a + 2 = 0$
$-2ah + 8a + 1 = 0$

I'm going to multiply the top equation by 2 and add the bottom equation to it:
$2ah + 2a + 4 = 0$
$-2ah + 8a + 1 = 0$
--------------------------------
$10a + 5 = 0$

Thus $a = -\frac{5}{10} = -\frac{1}{2}$

So
$ah + a + 2 = 0$
Becomes
$-\frac{1}{2}h + -\frac{1}{2} + 2 = 0$

$-h - 1 + 4 = 0$

$h = 3$

Finally:
$ah^2 + k = 8$

$-\frac{1}{2}(3)^2 + k = 8$

$-\frac{9}{2} + k = 8$

$-9 + 2k = 16$

$2k = 25$

$k = \frac{25}{2}$

So
$y = -\frac{1}{2}(x - 3)^2 + \frac{25}{2}$

-Dan

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