## [SPLIT] finding vertex, direction from y=ax^2+bx+c

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
pistolpete
Posts: 13
Joined: Wed May 20, 2009 1:07 am
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### [SPLIT] finding vertex, direction from y=ax^2+bx+c

My next section is a bit different and once again I have no clue what to do. In the book it says :

"Find the coordinates of the vertex then decide if the parabola has a minimum or maximum value."

Ex.1
f(x) = x^2 - 4x + 4

Ex. 2
f(x) = -x^2 - 8x - 16

Can you tell me what to do here?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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The formula for the vertex is:

$h\, =\, \frac{-b}{2a}$

...and then f(h) = k. This gives you the coordinates, (h, k). The sign of "a" tells you the direction: for a > 0, the vertex is the minimum because the parabola opens upward; for a < 0, the vertex is the maximum because the parabola opens downward.

pistolpete
Posts: 13
Joined: Wed May 20, 2009 1:07 am
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### Re: [SPLIT] finding vertex, direction from y=ax^2+bx+c

OKay Thanks You once again I got that now too.
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