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

factorise, and then one of them must equal 0, and there ur aswer. But im not sure that was ur question.

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

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

you need to find the three points where those two curves meet.

 

The functions only intersect twice.

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)

Ahhh my fault! sorry wrote the wrong function. y=x+1

excuse me

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)

True. Lets say that a=1.

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

Lovely!! How do you come about the formula to use?

I'm still not sure how to find c.

36-26-38

 

 

probably not at all helpfull, but a nice curve ratio, non the less

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

Aw cheers! Thanks a lot!