kingjewel1 Posted January 7, 2005 Share Posted January 7, 2005 ax^2+b^x+c is the standard function of a quadratic. If i know that a quadratic with this format passes though points A(-3,-2) and B(2,3). How would i go about working out a, b and c? This's got me stumped. Thanks Link to comment Share on other sites More sharing options...
ed84c Posted January 7, 2005 Share Posted January 7, 2005 factorise, and then one of them must equal 0, and there ur aswer. But im not sure that was ur question. Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 But i can't factorise because i don't know where it cuts the x axis. Maybe this helps: the curve passes through where y=6/x and y=x+1meet. I set them equal that's where i found the points A and B. Edit function is y=x+1 Link to comment Share on other sites More sharing options...
matt grime Posted January 7, 2005 Share Posted January 7, 2005 you need to find the three points where those two curves meet. then you know three points that satisfy y=ax^2+bx+c, and thus you can find a,b,c Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 you need to find the three points where those two curves meet. The functions only intersect twice. Link to comment Share on other sites More sharing options...
matt grime Posted January 7, 2005 Share Posted January 7, 2005 please forgive me if i suggest that you go away and check that. since a quick inspection tells me that the points of intersection you claim to have foudn for the two curves aren't correct either. 6/x = 2x-x^2 and you're claiming (-3, -2) is one of them.... let's check lhs is even, rhs is odd.... nope, that working. and (2,3), well, lhs is 3 at x=2. granted, but the rhs is zero when x=2 you need to solve x^3-2x+6=0 which has 3 roots (possibly with repeats, but I doubt it considering its derivative) Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 Ahhh my fault! sorry wrote the wrong function. y=x+1 excuse me Link to comment Share on other sites More sharing options...
The Rebel Posted January 7, 2005 Share Posted January 7, 2005 ax^2+b^x+c is the standard function of a quadratic.If i know that a quadratic with this format passes though points A(-3' date='-2)[/b'] and B(2,3). How would i go about working out a, b and c? This's got me stumped. Thanks Wouldn't you need three points to find three variables. For example the two points (0, 12) & (2, 32) satisfy [math]x^2+8x+12[/math] and also [math]3x^2+4x+12[/math] at least. Two points wouldn't even tell which way up the quadratic was (i.e. the x^2 coefficient) Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 True. Lets say that a=1. Link to comment Share on other sites More sharing options...
The Rebel Posted January 7, 2005 Share Posted January 7, 2005 True. Lets say that a=1. In that case for [math](x_1, y_1)[/math] and [math](x_2, y_2)[/math], [math]y_2-y_1 = x_2^2-x_1^2 + b(x_2-x_1)[/math] which equates to [math](y_2-y_1 - x_2^2-x_1^2) / (x_2-x_1) = b[/math] Then once b is found plug in the values to find c Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 Lovely!! How do you come about the formula to use? I'm still not sure how to find c. Link to comment Share on other sites More sharing options...
YT2095 Posted January 7, 2005 Share Posted January 7, 2005 36-26-38 probably not at all helpfull, but a nice curve ratio, non the less Link to comment Share on other sites More sharing options...
The Rebel Posted January 7, 2005 Share Posted January 7, 2005 Lovely!! How do you come about the formula to use?I'm still not sure how to find c. It was simply a little bit of algebra with simultaneous equations. You already had the formula y=ax^2+b^x+c. By imaging two points we had the formulae: y1 = ax1^2 + bx1 + c and y2 = ax2^2 + bx2 + c. We can cancel c by subtracting y1 from y2. To give us a formula with variables a and b. y2-y1 = a(x2^2-x1^2) + b(x2-x1) Because you stated a=1' date=' we can rearrange for b in terms of x1, y1, x2 and y2. That's how I came up with the formula. [b']b = (y2-y1 - (x2^2-x1^2)) / (x2-x1)[/b] As for c. . . Once you have found what b is you simply put the value of b into the formula along with say x1 and y1 to find c. e.g. y1 = x1^2 + bx1 + c => c = y1 - x1^2 - bx1 Link to comment Share on other sites More sharing options...
kingjewel1 Posted January 7, 2005 Author Share Posted January 7, 2005 Aw cheers! Thanks a lot! Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now