# 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)?

Jun 2, 2015

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

$f \left(x\right) = a \left(x + 2\right) \left(x - 7\right)$ for some constant $a$

Now we know that $f \left(2\right) = 9$, so:

$9 = f \left(2\right) = a \left(2 + 2\right) \left(2 - 7\right) = - 20 a$

Divide both sides by $- 20$ to get:

$a = - \frac{9}{20}$

So

$f \left(x\right) = - \frac{9}{20} \left(x + 2\right) \left(x - 7\right) = - \frac{9}{20} \left({x}^{2} - 5 x - 14\right)$

But there is a problem: $\left(2 , 9\right)$ is not the vertex of this quadratic. The vertex will have $x = \frac{5}{2} = \frac{- 2 + 7}{2}$ where

$f \left(\frac{5}{2}\right) = - \frac{9}{20} \left(\frac{5}{2} + 2\right) \left(\frac{5}{2} - 7\right)$

$= - \frac{9}{20} \left(\frac{5}{2} + \frac{4}{2}\right) \left(\frac{5}{2} - \frac{14}{2}\right)$

$= - \frac{9}{20} \cdot \frac{9}{2} \cdot - \frac{9}{2}$

$= \frac{729}{80} = 9.1125$

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, $\left(- 2 , 0\right)$ instead of $\left(- 3 , 0\right)$; or $\left(- 7 , 0\right)$ instead of $\left(- 6 , 0\right)$.
(2) The point $\left(9 , 2\right)$ 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.