Finding a quadratic equation of a function

[silent]

For this problem, I have to find the quadratic function ex. f(x) = quadratic equation.
I was given that the zeros of the function are 1 and 3 (so when I plug in 1 or 3 it equals 0)
It was also given the range is y >= -4 so I know the range is all numbers greater of equal to -4 which means -4 is the y coordinate of the vertex of the parabola (Am I correct?)

I have no idea how to proceed from here though..

 

[silent]

(x - 1)(x - 3) = 0
x^2 - 4x + 3 = 0


y = x^2 - 4x + 3
Try y = -4 and y = -5
Do you see why y=>-4 ?

Not really...I see how you find the quadratic function, but when I graph it I get a vertex of (2,-1) for the parabola and i know the domain is all real numbers, but this function would make the range >= -1 ...am I thinking right?

 

[DrPhil]

Not really...I see how you find the quadratic function, but when I graph it I get a vertex of (2,-1) for the parabola and i know the domain is all real numbers, but this function would make the range >= -1 ...am I thinking right?

Yes. The domain refers to values of x for which the function is defined, and that is "all real numbers." The range is the set of y that occur for all x in the domain. The range is exactly what you stated.

 

[lookagain]

For this problem, I have to find the quadratic function ex. f(x) = quadratic equation.

I was given that the zeros of the function are 1 and 3 (so when I [substitute] in 1 or 3 it equals 0)

It was also given the range is y >= -4 so I know the range is all numbers greater of equal to -4 which
means -4 is the y coordinate of the vertex of the parabola (Am I correct?)\(\displaystyle \ \ \ \)The y-coordinate is -4, yes.

I have no idea how to proceed from here though..

User silent, so there still isn't an answer for you in is thread so far.

The x-value the vertex will be the average of the zeroes of this particular quadratic, which is 2.

The vertex is (2, -4).

For some positive real-valued variable, a, the function at this stage is

f(x) = a(x - 1)(x - 3)


Substitute in the coordinates of the vertex and solve for a:

-4 = a(2 - 1)(2 - 3)

-4 = a(1)(-1)

-4 = -a

4 = a

a = 4


So, the quadratic function that satisfies those conditions given in the original statement of the problem is

f(x) = 4(x - 1)(x - 3)


\(\displaystyle f(x) \ = \ 4(x^2 - 4x + 3) \)


\(\displaystyle \boxed{ \ f(x) \ = \ 4x^2 - 16x + 12 \ } \ \ \ \ \ \ \ \ \)You see when you substitute the x-value of the vertex (x = 2), you'll get f(x) = -4.

 

[stapel]

I see how you find the quadratic function, but when I graph it I get a vertex of (2,-1)....

The zeroes only give you the variable-containing factors. There could still be numerical-only factors, which the zeroes don't address (because they are unaffected by those numerical-only factors). In other words, the zeroes only give you a "family" of quadratics; to find "the" quadratic, they have to give you another piece of information, such as another point on the graph or, as in this case, the range of the function.

You can see some worked examples and explanations here. In general, given zeroes p and q, the factors are (x - p) and (x - q). But the quadratic will then be f(x) = a(x - p)(x - q), and "a" can be ANY value (other than zero). There are infinitely-many possible quadratics, if one is given only the zeroes. When given (in your case) the range limit, you then have enough information to find "a": the vertex is at the x-value halfway between the zeroes (you know this, from the graphing you've done), so you plug that value in for x, plug the range limit in for y, and then solve for "a".