How to find function with given roots and point

In summary, the Homework Statement is about finding 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
Eleeist
11
0

Homework Statement



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?
 
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  • #2
Substituting values in [itex]y=ax^2+bx+c[/itex] is a good approach. Can you tell us what you get if you do this?
 
  • #3
Well, I know c = 8, but I still don't know the values of a and b.

ax^2+bx+8=y.
 
  • #4
-2 and 2 are x-intercepts, what does that mean?? What points lie on the parabola??
 
  • #5
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.
 
  • #6
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.
 
  • #7
Ok, got it:

-2x^2+8=y

Thanks.
 
  • #8
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.
 
  • #9
And what if the point which the function crosses is not y-intercept, eg (4,18)?
 
  • #10
Eleeist said:
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.
 
  • #11
But then I don't have b and a.
 
Last edited:
  • #12
Eleeist said:
And what if the point which the function crosses is not y-intercept, eg (4,18)?

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

Eleeist said:
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).
 
  • #13
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.
 
  • #14
Eleeist said:
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.
 
  • #15
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)?
 
  • #16
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.
 
  • #17
I understand that part. But how did the c get before brackets?
 
  • #18
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?
 
  • #19
eumyang said:
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.
 
  • #20
eumyang said:
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 said:
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.
 

1. What is a function?

A function is a mathematical relationship between inputs and outputs. It takes in one or more inputs and produces a unique output for each input.

2. How do I find the function with given roots and point?

To find the function, you will need to use the information provided about the roots and point to create an equation. You can use the general form of a polynomial function, which is y = a(x-r1)(x-r2)...(x-rn), where a is the leading coefficient and r1, r2, ..., rn are the roots. Plug in the coordinates of the given point to solve for the value of a. This will give you the function in the form of y = ax(x-r1)(x-r2)...(x-rn).

3. Can a function have more than one root?

Yes, a function can have multiple roots. The number of roots a function has is equal to its degree, or the highest exponent in the equation. For example, a quadratic function can have up to two roots, while a cubic function can have up to three roots.

4. What if the given point is not on the function?

If the given point is not on the function, then the function does not pass through that point. In this case, you can use the information about the roots and the form of the function to determine the behavior of the function near the given point. You can also use the point-slope form to find the equation of the tangent line to the function at that point.

5. Is there a way to check if my function is correct?

Yes, there are a few ways to check if your function is correct. First, you can graph the function and see if it passes through the given point and intersects the x-axis at the given roots. Another way is to plug in the coordinates of the given point and the roots into the function and see if they satisfy the equation. You can also use a graphing calculator or online function plotter to verify your function.

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