# Porabola and Vertex

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• Dec 10th 2012, 02:31 PM
maiden129
Porabola and Vertex
For f(x) = -2x^2 - 8x-2

a. Determine whether the parabola opens up or down

b. Find the vertex

how can determine whether the parabola opens up or down and find the vertex? Thanks in advance.
• Dec 10th 2012, 02:46 PM
skeeter
Re: Porabola and Vertex
$\displaystyle f(x) = ax^2 + bx + c$

if the leading coefficient, $\displaystyle a$ , of the quadratic is positive, the parabola opens upward

if the leading coefficient, $\displaystyle a$ , of the quadratic is negative, the parabola opens downward

the x-value of the vertex is $\displaystyle x = \frac{-b}{2a}$
• Dec 10th 2012, 03:04 PM
sakonpure6
Re: Porabola and Vertex
It opens down because a is negative as skeeter mentioned as as for the vertex, I also can assume you have not learned how to do vertex form? because the equation you provided does not factor.
• Dec 10th 2012, 03:10 PM
skeeter
Re: Porabola and Vertex
Quote:

Originally Posted by sakonpure6
It opens down because a is negative as skeeter mentioned as as for the vertex, I also can assume you have not learned how to do vertex form? because the equation you provided does not factor.

the quadratic does not need to be factorable to change it to vertex form.
• Dec 10th 2012, 03:11 PM
sakonpure6
Re: Porabola and Vertex
Quote:

Originally Posted by skeeter
the quadratic does not need to be factorable to change it to vertex form.

I know but I think he is new to this, other wise he would have made a topic on vertex form