# Thread: Determining an expression for a function

1. ## Determining an expression for a function

I can't figure this out for the life of me. Help is much appreciated.

The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.

Thanks!

2. Originally Posted by ocmisssunshine
The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.
Here we want to make a quadratic equation.

So x must = either -1 or 3 for y to be 0 and to intercept the x-axis

Therefore (x+1)(x-3) = 0
expanding gives you... x^2 - 3x + x - 3 = 0 where x <= 4

the graph is as seen below

3. Oh thanks!!! That helps alot!!

4. Originally Posted by ocmisssunshine
I can't figure this out for the life of me. Help is much appreciated.

The graph of a quadratic function (a parabola) has x-intercepts -1 and 3 and a range consisting of all numbers less than or equal to 4. Determine an expression for the function.

Thanks!
(We were sent home due to heavy rain, so I can help you now.)

The parabola has two x-intercepts. That means it is a "vertical" parabola, or a parabola whose axis of symmetry is vertical, or a parabola that opens upward or downward.

The range is all numbers less than or equal to 4. That means the parabola is below, or on, the y=4 horizontal line. That means further that the vertical parabola opens downward.
The "equal to 4" means the parabola touches the y=4 at one point only. That means further that the the vertex is at y=4.

A standard form of a vertical parabola that opens downward is
(y-k) = -a(x-h)^2 ---------(i)
where
(h,k) is the vertex
"a" is any rational number.

The two x-intercepts -1 and 3 are the points (-1,0) and (3,0) respectively.

For point (-1,0),
(y-k) = -a(x-h)^2 ---------(i)
(0-k) = -a(-1-h)^2
-k = -a(1+h)^2
k = a(1+h)^2 ---------------(ii)

For point (3,0),
(y-k) = -a(x-h)^2 ---------(i)
(0-k) = -a(3-h)^2
-k = -a(3-h)^2
k = a(3-h)^2 ---------------(iii)

The k in (ii) equals the k in (iii),
a(1+h)^2 = a(3-h)^2
(1+h)^2 =(3-h)^2
1+h = 3-h
h+h = 3-1
2h = 2
h = 1 ---------------**

We already know that k = 4, so (h,k) = (1,4).
Plug those into (i), using the x-intercept point (-1,0)
(y-k) = -a(x-h)^2 ---------(i)
(0-4) = -a(-1-1)^2
-4 = -4a
a = 1 ------------**

Therefore, the equation of the parabola is
(y-k) = -a(x-h)^2 ---------(i)
y -4 = -1(x-1)^2
y -4 = -(x^2 -2x +1)
y = -x^2 +2x -1 +4
y = -x^2 +2x +3 -------------------answer.