What is the formula for the family of quadratic functions passing through (3,0)?

[wellY--3]

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?

 

[theperthvan]

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.

 

[ssd]

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.

 

[wellY--3]

what??

and put in the original what

 

[wellY--3]

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 let's 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?

 

[HallsofIvy]

The point is that any "quadratic function" can be written in the form
y= f(x)= ax²+ bx+ c. You want to write a formula that describes the "family" (i.e. set) of all those that pass through (3,0)- that is, all those for which y= 0 when x= 3. Putting y= 0 and x= 3 into that original formula,
0= 9a+ 3b+ c so c= -(9a+3b). The answer to the question is that the family of all quadratic functions that pass through (3, 0) are those of the form f(x)= ax²+ bx- (9a+ 3b).

(That's one way to write the answer: we could also, of course, have solved 9a+ 3b+ c= 0 for a, in terms of b and c, or for b, in terms of a and c, and replaced that parameter instead of c.)