Question

How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (-2,0) & (7,0)?

Answer 1

Given the intercepts, the quadratic must be of the form:

`f(x) = a(x+2)(x-7)` for some constant `a`

Now we know that `f(2) = 9`, so:

`9 = f(2) = a(2+2)(2-7) = -20a`

Divide both sides by `-20` to get:

`a = -9/20`

So

`f(x) = -9/20(x+2)(x-7) = -9/20(x^2-5x-14)`

But there is a problem: `(2, 9)` is not the vertex of this quadratic. The vertex will have `x = 5/2 = (-2+7)/2` where

`f(5/2) = -9/20(5/2+2)(5/2-7)`

`=-9/20(5/2+4/2)(5/2-14/2)`

`=-9/20*9/2*-9/2`

`=729/80 = 9.1125`

So the problem as stated seems to be incorrect in one of the following ways:
(1) One of the intercepts was given incorrectly. For example, `(-2, 0)` instead of `(-3, 0)`; or `(-7, 0)` instead of `(-6, 0)`.
(2) The point `(9, 2)` is just a point through which the quadratic passes - not necessarily a vertex.
(3) The parabola is tilted from the vertical, described by a 2nd order equation in `x` and `y`. That would not express `y` as a quadratic function of `x` or vice-versa.