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.