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

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Answer 1 · 2016-12-31T15:40:22

Please see the explanation.

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

$y = k (x + 1) (x - 3)$ $y = k(x + 1)(x - 3)$

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

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

I will multiply the factors:

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

I will chose $k = 1$ $k = 1$

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

Now I will choose $k = - 1$ $k = -1$

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

Equation [1] opens upward and equation [2] opens downward; both have the same intercepts, $(- 1, 0) and (3, 0)$ $(-1,0) and (3,0)$