Graphing quadratic equations.

Okay, lets start from ground 0 of the whole process.

What do I do with my equation? y= ax(squared)+bx+c ?

Do I factor it or not?

If not, is it what I put in my table of values and if so what do I put in my table of values?

 

y=9x(squared)-36x+35 <--- so for this replace every X with chosen value, and solve for Y... Do it with 5 values for X and you have your table...

 

y = 9(-10)^2 - 36(-10) + 35 etc....

 

Or actually maybe you have to do the factor thing or something, then simplify it, and then solve for Y...

 

Damnit now I'm confused...

 

Okay, 2025+180+35 = 2240 (using -5 as my value for x)
729+108+35 = 872 (using -3 as my value for x)
0+36+35 = 71 (using 1- as my value for x)
324-72+35 = 287 (using 2 as my value for x)

Does this seem right so far?

 

ok, to solve for y you factor the eqyation then each of the factor will equal zero....you then re-arrange each of the facotrs seperatly so that you can solve for x.......(during the whlole time y=0) until you re-arrange the factor...this makes sense in my head...

 

that would be right to make a table and to graph after...

 

That sounds familiar to me but doesn't ring a bell as far as making sense, but I wouldn't take my word for sure. I was left way back there, lol. Slightly getting back on track. (The thing is, I understand it more in writing than I do as numbers and directions shown to me. But it's easier for me to teach to others doing it as examples.)

 

How would you go about doing that?

 

Great, cause thats exactly what I'm going for.

 

Well, my factors for this particular question are (x+5)(x-2).
How would I rearrange that?

 

x=-5 and x=-2

they are like 2 mini-equations...each equals zero and those are your solutions for x

thats all i can do for now...good luck

 

Burbot is right. Those two numbers then -5 abd -2 will be the points on the graph that the parabola crosses the x-axis

 

Damn all the things online make it more complicated than it is...
What you have to do is simplify it until its something like Y = +/-X^2 I believe, then just make the chart...

 

(x+5)(x-2) = x^2 + 3
so Y + X^2 + 3
so enter the X values and you got your chart!

 

So:
Y = (-10)^2 + 3
Y = (-8)^2 + 3
etc. from X = -10 to X = 10

 

To find the point were the graph crosses the y-axis. set x=0 so it will look like this y=((0)+5)((0)-2) goes to y=5*-2 goes to y=-10 so the graph crosses the y axis at -10.

 

Well, obviously 0 is included between -10 and 10...

 

souless chaos, I am not sure where you are getting the (x+5)(x-2)=(x+3) that doesn't quite work out. Since they are in parenthesis they must be multiplied. (x+5)(x-2)=(x^2+3x-10), but the factored form is easier to use.