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:
$\displaystyle y = a(x - h)^2 + k$
You have zeros at (-2, 0) and (8, 0) and a y-intecept of (0, 8).
Plugging these points into the equation individually:
$\displaystyle 0 = a(-2 - h)^2 + k$
$\displaystyle 0 = a(8 - h)^2 + k$
$\displaystyle 8 = a(0 - h)^2 + k$
Or:
$\displaystyle ah^2 + 4ah + 4a + k = 0$
$\displaystyle ah^2 - 16ah + 64a + k = 0$
$\displaystyle ah^2 + k = 8$
Probably the simplest thing to do here would be to take the last equation:
$\displaystyle ah^2 = 8 - k$ and sub it into the top two equations.
$\displaystyle (8 - k) + 4ah + 4a + k = 0$ ==> $\displaystyle 4ah + 4a + 8 = 0$
$\displaystyle (8 - k) - 16ah + 64a + k = 0$ ==> $\displaystyle -16ah + 64a + 8 = 0$
These turn into:
$\displaystyle ah + a + 2 = 0$
$\displaystyle -2ah + 8a + 1 = 0$
I'm going to multiply the top equation by 2 and add the bottom equation to it:
$\displaystyle 2ah + 2a + 4 = 0$
$\displaystyle -2ah + 8a + 1 = 0$
--------------------------------
$\displaystyle 10a + 5 = 0$
Thus $\displaystyle a = -\frac{5}{10} = -\frac{1}{2}$
So
$\displaystyle ah + a + 2 = 0$
Becomes
$\displaystyle -\frac{1}{2}h + -\frac{1}{2} + 2 = 0$
$\displaystyle -h - 1 + 4 = 0$
$\displaystyle h = 3$
Finally:
$\displaystyle ah^2 + k = 8$
$\displaystyle -\frac{1}{2}(3)^2 + k = 8$
$\displaystyle -\frac{9}{2} + k = 8$
$\displaystyle -9 + 2k = 16$
$\displaystyle 2k = 25$
$\displaystyle k = \frac{25}{2}$
So
$\displaystyle y = -\frac{1}{2}(x - 3)^2 + \frac{25}{2}$
-Dan