How do you use factoring to find the zero's of equation? And finding the vertex and axis of symmetry?

To find the zero's of the equations you have to factor them into linear equations, which means equations that have no degree. So I'm uming you know how to factor, but if you don't you need to find factors of the constant (numbers that can divide into it evenly) that equal the coefficient of just x. So what are the factors of -24? (-)24 and (-)1 ( can be limited out because there is no way these number can equal 2 when added or subtracted), (-)6 and (-)4 ( this is it because 6 + -4 is 2 but lets check the others just in case), (-)12 and (-)2 ( this can't be it for the same reason as the first pair), and (-)8 and (-)3 ( while it seems like this could be it, they cant equal 2). So now that you know the two numbers are -4 and 6 you can properly factor this to get (x-4)(x+6) because the degree is 2 (x^2) and that means that it is x+/- whatever times x +/- whatever. Now the zeros of the equation are the numbers that cause each individual factor(x-4 and x+6 are factors of the original.) of the equation to equal 0 if they are substituted for x. So what -4 is 0? 4, right? So one zero is 4, and if you do the same thing for X+6, so the other zero is -6. So if you do the same thing for y=x^2-7x-28, you might get confused. Since no factor pairs of -28 equal -7, this is a prime function. What I mean by prime is that it cannot be factored and therefore you cannot find its zeros the same way. So to find the zeros of this function, you will have to use the quadratic formula. The quadratic formula is used to solve functions that cannot be factored. So to do this you must know that the coefficient (the number next to the variable) of x^2 is the same as A, the coefficient of x (-7) is B, and the constant (-28) is C. Just a warning, the quadratic formula will require a calculator. The formula is (-B +/- (radical) B^2 - 4AC)/2A. So you have to use this equation twice: one with -B + the rest and one with -B - the rest of the equation. When you do this you get something like 9.8443 and -2.84423 as your answers, which are also the zeroes of the function (those numbers are rounded). So I hope that clears up the zeros of functions. Now, let's move onto the vertex. When you have the equation/function in standard form, you do (-B)/2A to get the x coordinate of the vertex of the equation. So for y=3(x-3)^2+4, you first have to expand (x-3)^2 for the purpose of simplicity. So (x-3)^2 is the same as (X-3) x (X-3). So you can use FOIL (first, outer, inner, last; if you don't know what FOIL is, then see a href="http://www.mathwords.com/f/foil_method.htm" rel="nofollow"http://www.mathwords.com/f/foil_method.h…/a ). So you get y=3(X^2-6x+9)+4. So now you distribute the 3 to get y=(3x^2-18x+27)+4. Then add 4 to get y=3x^2-18x+31. Now, you can use the (-B)/(2A) to get the x coordinate of the vertex. So plug in B and A to get -(-18)/(2 x 3). - (-18) is the same as just 18 so you get 18/6, which is 3. So the x coordinate of the vertex is 3. Now you plug 3 in to get the y coordinate. So y=3(3-3)^2+4, which is y=3(0)^2+4 then its 3(0)+4, then 0+4 and then just y=4. So the vertex's coordinates are (3,4). Now the axis of symmetry is much easier. It is just the x coordinate of the vertex! So the axis of symmetry is x= the x of the vertex, so for this equation, the axis of symmetry is x=3. Now using what I just old you, the vertex of y=x^2+3x-4 is (-3/2, -25/4).