How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (-1,0) & (5,0)?

Jun 2, 2015

From the $x$ intercepts, we can deduce:

$f \left(x\right) = a \left(x + 1\right) \left(x - 5\right)$ for some constant $a$

Then from the vertex point we get:

$9 = f \left(2\right) = a \left(2 + 1\right) \left(2 - 5\right) = - 9 a$

So $a = - 1$

$f \left(x\right) = - \left(x + 1\right) \left(x - 5\right) = - \left({x}^{2} - 4 x - 5\right)$

$= - {x}^{2} + 4 x + 5$