Question

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)?

Answer 1

Many possible answers.
One such: `x^2+7/2x+3/2` and `-x^2-7/2x-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(x) = (x+3)(x+1/2)`
`f(x) = x^2 +7/2x +3/2`

Since this function `f(x)` 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(x) = -(x+3)(x+1/2)`
`g(x) = -x^2 -7/2x-3/2`

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