A family of quadratic functions passes through the points (3,0). Find the family of quadratic functions

err i have no idea hwo to do it except substituting those values in ... 0=9a+3b+c

what does it meant the family of quadratic functions?

And then find a in terms of b and c
b in terms of a and c
c in terms of a and b
then put them back into the original quadratic.

And then find a in terms of b and c
b in terms of a and c
c in terms of a and b
then put them back into the original quadratic.

From two equations, in general, you at most can eliminate only one of the unknowns. So from the original y=ax^2 +bx +c, you can remove only one of a or b or c.
Write, a in terms of b and c (or, b in terms of a and c; or, c in terms of a and b) and put in the original.... that is your family of equations for any choice of the existing two parameters.

what??

and put in the original what

so c =-12x
is that right?

uart
No, c = -(9a + 3b).

So the family is,

$$y(x) = a x^2 + b x -(9a + 3b)$$

The one given condition only lets you eliminate one unknown parameter. So you end up with a quadratic function that still has two free parameters, that's why it's referred to as a "family", there's lot of 'em. Get it?

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HallsofIvy