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

Dec 31, 2016

#### Explanation:

Because the quadratic function is zero, when $x = - 1 \mathmr{and} x = 3$, it will have the factors:

$y = k \left(x + 1\right) \left(x - 3\right)$

where k is an unknown constant that one can use to force the quadratic to pass through a point with a non-zero y coordinate.

If $k > 0$, then the quadratic opens upward.

If $k < 0$, then the quadratic opens downward.

I will multiply the factors:

$y = k \left({x}^{2} - 2 x - 3\right)$

I will chose $k = 1$

$y = {x}^{2} - 2 x - 3 \text{ [1]}$

Now I will choose $k = - 1$

$y = - {x}^{2} + 2 x + 3 \text{ [2]}$

Equation [1] opens upward and equation [2] opens downward; both have the same intercepts, $\left(- 1 , 0\right) \mathmr{and} \left(3 , 0\right)$