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)?

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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)?

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Answer 1 · 2017-10-01T05:35:04

Many possible answers.
One such: $x^{2} + \frac{7}{2} x + \frac{3}{2}$ $x^2+7/2x+3/2$ and $- x^{2} - \frac{7}{2} x - \frac{3}{2}$ $-x^2-7/2x-3/2$

Explanation:

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

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

Since this function $f (x)$ $f(x)$ has a positive $x^{2}$ $x^2$ 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:

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

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

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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)?

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