# How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (-3,0), (-1/2,0)?

Oct 1, 2017

One such: ${x}^{2} + \frac{7}{2} x + \frac{3}{2}$ and $- {x}^{2} - \frac{7}{2} x - \frac{3}{2}$

#### Explanation:

We are told that one quadratic opens up and the other opens down. In addition we have two $x$-intercepts. Recall that $x$-intercepts are synonyms for zeroes, and those can be written as factors, leading us to a way to derive an equation for the function:

$f \left(x\right) = \left(x + 3\right) \left(x + \frac{1}{2}\right)$
$f \left(x\right) = {x}^{2} + \frac{7}{2} x + \frac{3}{2}$

Since this function $f \left(x\right)$ has a positive ${x}^{2}$ term, this parabola opens upward.

To get another function that opens downward, the easiest solution is to simply put a negative in front of the factors:

$g \left(x\right) = - \left(x + 3\right) \left(x + \frac{1}{2}\right)$
$g \left(x\right) = - {x}^{2} - \frac{7}{2} x - \frac{3}{2}$

graph{(x^2+7/2x+3/2-y)(-x^2-7/2x-3/2-y)=0 [-10, 10, -5, 5]}