Find the zeros y=x^2+x-12 and the equation of the axis symmetry

Question

anonymous · Answer

again?  it is still 
$$x=-\frac{b}{2a}$$ and y is what you get by substitution

anonymous · Answer

I dont understand .

anonymous · Answer

For quadratic of form x^2 + bx + c , u can try to find 2 numbers p and q that add up to b (1 in this case)  and multiply together to give c (-12 in this case). Then the factors are (x+p)(x+q). 
Often works with made up problems, else use quadratic formula.

You can differentiate for the turning point, do you know how to?

anonymous · Answer

No .

anonymous · Answer

General form of quadratic is ax^2 + bx + c so you can use satellite73 formula for x coordinate of axis of symmetry, then substitute this value in the equation to get the y-coordinate.

The zeros u can get using the method I said above.

anonymous · Answer

@estudier this has nothing to do with differentiation, nor the quadratic formula since he/she is asking for vertex.

anonymous · Answer

oops my fault.  zeros are asked for as well.  i apologize

anonymous · Answer

- Im a SHE . My name is Lyssah as in LISA . Juss spelled differently .

anonymous · Answer

For me it is just easy to say 2x +1 = 0...:-)

anonymous · Answer

@lyssah
you have something that looks like 
$$y=ax^2+bx+c$$ vertex is 
$$x=-\frac{b}{2a}$$ and y you get by substitution. i do no presume anyone is any specific gender on the internet

anonymous · Answer

I was juss clarifying .

anonymous · Answer

@estydier you are using calc which these problems probably precede.  just use 
$$-\frac{b}{2a}$$ that is it

anonymous · Answer

I did ask if she knew, when she said no, I referred her to your formula and explained what a b c meant.