Family of quadratic functions?

[mathman100]

This question has been killing me for days, :
Give a family of quadratic functions of the form y=ax^2+bx+c, that passes through the following points:
(1,1) and (2,0)
I see how we can find the family, but how do we find the specific functions that pass through both those points? I tried making two separate equations (1 for each point, like(1,1) I would sub in for x and y) but that didn't give me anything useful. Even if I use elimination for the 2 equations, I don't know where it will get me...what should i do?

 

[Data]

I tried making two separate equations (1 for each point, like(1,1) I would sub in for x and y) but that didn't give me anything useful.

Of course it does. You get the two equations

$$1 = a + b + c$$

and

$$0 = 4a + 2b + c,$$

which can be solved to get $b = -1 - 3a$ and $c = 2(a+1)$ with no conditions on $a$. So what's your family of functions?

 

[mathman100]

Really? Well first of all I put down the wrong 2 points (aorry, my fault) they should be (1,) and (-1,-2). I got 0=a+b+c and -2=a-b+c. I solved to get b=-1 and i think c=-1-a. I don't know how to continue!

 

[mathman100]

woops, points= (1,0) and (-1,-2)

 

[mathman100]

does anyone know?

 

[Data]

well, if b=-1 and c=-1-a in ax^2 + bx + c then what do the polynomials that go through those points look like?

 

[mathman100]

i don't know, now I'm lost do you mean like a parabola? a linear line with a slope? how do i define the family? why can't i just sub in the known values of b and c:
ax^2-x-1-a?

 

[Data]

you can! that is exactly the "family" it's looking for: all the quadratics with the form $a^2 + x - 1 - a$ for nonzero constants $a$ (you actually made an error earlier. You should have found $b=1$, not $b=-1$).

 

[murshid_islam]

you can! that is exactly the "family" it's looking for: all the quadratics with the form $a^2 + x - 1 - a$ for nonzero constants $a$

should it be $$a^2 + x - 1 - a$$? or $$ax^2 + x - 1 - a$$?

 

[Data]

$ax^2 + x - 1 -a.$ I'm not very good at typing!