How to find function with given roots and point

[Eleeist]

1. The problem statement, all variables and given/known data

Find the quadratic function which has x-intercepts -2 and 2 and passes through the point (0,8).

I am struggling with solving this one. I tried to substitute values into ax^2+bx+c=y but there are too many unknowns.

Can anyone give some tips how to solve it?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[micromass]

Substituting values in y=ax^2+bx+c is a good approach. Can you tell us what you get if you do this?

[Eleeist]

Well, I know c = 8, but I still don't know the values of a and b.

ax^2+bx+8=y.

[micromass]

-2 and 2 are x-intercepts, what does that mean?? What points lie on the parabola??

[Eleeist]

It means that they cross the x axis. And where they cross y=0... But then I get this:

a(2)^2+2b+8=0

I still have two unknowns.

[micromass]

Yes, so 2 is an x-intercept thus

4a+2b+8=0

Also -2 is an x-intercept, thus: ...


You'll end up with a system of 2 equations and 2 unknowns.

[Eleeist]

Ok, got it:

-2x^2+8=y

Thanks.

[LCKurtz]

You also might have noticed that, given the roots -2 and 2 your equation must have been $y = c(x-2)(x+2) = c(x^2-4)$ which would have had only one unknown.

[Eleeist]

And what if the point which the function crosses is not y-intercept, eg (4,18)?

[LCKurtz]

And what if the point which the function crosses is not y-intercept, eg (4,18)?

Are you addressing that question to me? What difference does that make? Put in x = 4, y = 18 and solve for c.

[Eleeist]

But then I don't have b and a.

[LCKurtz]

And what if the point which the function crosses is not y-intercept, eg (4,18)?

Are you addressing that question to me? What difference does that make? Put in x = 4, y = 18 and solve for c.

But then I don't have c.

Huh?? If you put x=4, y = 18 into y = c(x2-4) you get 18 = c(16-4).

[Eleeist]

Sorry, I meant b and a and was referring to the equation in form ax^2+bx+c=y. How would I go about using this form to solve the problem?

Also, could you please explain in more detail this:

y=c(x−2)(x+2)

I think I am seeing this form for the first time, but I might be wrong.

[LCKurtz]

Sorry, I meant b and a and was referring to the equation in form ax^2+bx+c=y. How would I go about using this form to solve the problem?

Also, could you please explain in more detail this:

y=c(x−2)(x+2)

I think I am seeing this form for the first time, but I might be wrong.

Presumably you know the factor theorem: If r is a root of a polynomial, then (x-r) is a factor. So your two given roots imply that (x-2) and (x+2) are factors. Two linear factors make a quadratic and the only question is a possible multiplier, or coefficient of x2. That is you you can deduce the form of the polynomial must be:

y=c(x−2)(x+2).

Then putting in your other point determines c. It is equivalent to how you solved the problem, but a little easier.

[Eleeist]

Yes, I know the factor theorem. But I cannot understand how you can deduce y=c(x-2)(x+2). Do you suggest that c is coefficient of x^2 (of course it is to)?

[eumyang]

You are given that 2 and -2 are the x-intercepts. That means that
(x - 2) and (x - (-2)) = (x + 2)
are factors of this quadratic.

So y has to equal
y = a(x - 2)(x + 2)
(I don't like using c -- to be consistent with the OP I'll use a)
Plug in the point (0, 8) and you'll have a.

[Eleeist]

I understand that part. But how did the c get before brackets?

[eumyang]

Say that 3 and 5 are the zeros of another quadratic. Then, without any further information, you could say that the quadratic is
y = (x - 3)(x - 5) = x2 - 8x + 15.

But that's not the only quadratic that would work. This one would also work:
y = 5x2 - 40x + 75
If you find the zeros of this quadratic, you will find that they will also be 3 and 5. This quadratic happens to be 5 times the first quadratic:
5x2 - 40x + 75 = 5(x2 - 8x + 15)

In general, given the zeros of a quadratic p and q (and nothing else), the quadratic would be
y = a(x - p)(x - q)

However, in the OP, a point is also given, which means that we need to find a specific value of a that would allow the quadratic to go through that point.


P.S. You realize that the c in y = c(x - 2)(x + 2) is not the same as the c in y = ax2 + bx + c, do you?

[Eleeist]

P.S. You realize that the c in y = c(x - 2)(x + 2) is not the same as the c in y = ax2 + bx + c, do you?

No. I actually realized that just now :). I thought you meant "c" as y-intercept. But now that you explained that it is rather "a", then I understand the logic.

Thanks for detailed explanation.

[LCKurtz]

P.S. You realize that the c in y = c(x - 2)(x + 2) is not the same as the c in y = ax2 + bx + c, do you?

No. I actually realized that just now :). I thought you meant "c" as y-intercept. But now that you explained that it is rather "a", then I understand the logic.

Thanks for detailed explanation.

Good job of catching that eumyang. I hadn't caught that his difficulty was with the fact I had used "c" instead of "a" in my explanation.