Math Is Fun Forum

Titus

Member

Registered: 2005-03-07

Posts: 10

Quadratic Functions

I have this question which asks;

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A family of quadratic functions just touches the x-axis at the point (3,0). Find the family of quadratic functions.

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What exactly is a family of quadratic equations and how do i find it?

Also, how would i go about finding one particular function from the family?

Thankyou.

Ansette

Member

Registered: 2006-02-19

Posts: 21

Re: Quadratic Functions

well.. a quadratic function is generally given in the form:

ax²+bx+c=0

where a,b,c are real constants.

The value of x is the point on the x-axis that the quadratic function touches it, if it touches once then there's one solution however it can cross it then come back up as it's a curve and thus there are 2 solutions.

Now, you know there's only one solution at (3,0).. e.g. x=3.

therefore the family logically is:

a(3²)+b(3)+c=0

therefore:

9a+3b+c=0

(or alternatively as c is whats called a dummy variable, you could simply write it using a new value and simplify it)

3a+b=d

hope this makes sense, i am very tired lol :s

francia

Guest

Re: Quadratic Functions

I have this question which asks;

--

A family of quadratic functions just touches the x-axis at the point (3,0). Find the family of quadratic functions.

--

What exactly is a family of quadratic equations and how do i find it?

Also, how would i go about finding one particular function from the family?

Thankyou.

Devantè

Real Member

Registered: 2006-07-14

Posts: 6,400

Re: Quadratic Functions

And your point in quoting that message was?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Quadratic Functions

That isn't the full solution. You're right that the quadratics that are part of that family must have the property that 9a+3b+c=0, but that only makes them have a point at (3,0). It doesn't force them to have a turning point there. y=x(x-3), for example, has a point at (3,0) but also goes past it.

Therefore you need to impose another restriction.

Differentiating the function gives you 2ax+b. Turning points occur when this is zero, and so you need 2a(3)+b, or 6a+b=0 to be true.

This can be rearranged to give -6a=b, which in turn can go into the first equation to make it 9a-18a+c=0. This means that c must equal 9a for the equation to be in the family.

Therefore, your final answer is that the family of quadratics take the form ax²-6ax+9a = 0, for all real values of a.

By factorising, this becomes a(x-3)²=0.

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Why did the vector cross the road?
It wanted to be normal.