# How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (-5/2,0), (2,0)?

Jan 15, 2017

$f {\left(x\right)}_{1} = 2 {x}^{2} - x - 10$
$f {\left(x\right)}_{2} = - 2 {x}^{2} + x + 10$

#### Explanation:

A quadratic function's opening can be determined by the sign of the term with degree 2. If the sign is positive, the quadratic opens upward. Otherwise, the quadratic opens downward. To get the quadratic equation opening downward, simply scale the trinomial by -1.

The roots of the quadratic equation are already given. We can obtain the "base" trinomial from the roots

$f \left(x\right) = 0 = \left(\pm 1\right) \left(x + \frac{5}{2}\right) \left(x - 2\right)$

$\implies f \left(x\right) = \left(\pm 1\right) \left({x}^{2} - \frac{x}{2} - 5\right)$

$\implies f \left(x\right) = \left(\pm 1\right) \left(2 {x}^{2} - x - 10\right)$

$\implies f {\left(x\right)}_{1} = 2 {x}^{2} - x - 10$

$\implies f {\left(x\right)}_{2} = - 2 {x}^{2} + x + 10$