Y=x^2-4x-12

Question

Y=x^2-4x-12

er.mohd.amir · Answer

here last term is -12 break it into two such number then subtraction is -4

er.mohd.amir · Answer

that is equal to middle term

mathmale · Answer

Please, would you include the instructions along with the expression x^2 - 4X - 12?

mathmale · Answer

Factor?  Find the zeros?  Graph it?  ??

mathmale · Answer

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anonymous · Answer

Thank you... How would you get the maximum/minimum of the parabola.  :)

calculusxy · Answer

My best suggestion is for you to graph it and find out the vertex of the parabola. If the parabola is going up, which I think it should because the $a$ terms is positive, then the parabola will be a minimum. However, if the parabola faces downwards, then it will be the maximum point of the parabola.

calculusxy · Answer

But you can figure out the vertex by finding the axis of symmetry and then plug it back to the equation to find the y-coordinate of the vertex. 

Formula for axis of symmetry:
$\large x = \frac{-b}{2a}$

mathmale · Answer

In the case of a parabola, finding the vertex is equivalent to finding the max. or the min.  
If you follow the advice given you by calculusxy (agove), you will end up with the x-coordinate of the vertex.  Substitute this value into the original function to find the y-coordinate.

Have you identified a max or a min?  To do so, determine the sign of the x^2 term.  If it's poistive, your parabola opens upward, and you have a minimum at the vertex.  If negative, your parabola opens downward and you have a max. at the vertex.

mathmale · Answer

Do you know some calculus already?  If so, finding the first and second derivative of the given equation will quickly tell you where your max / min is and whether the parabola opens up or down.