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)?

1 Answer
Jason K.
Oct 1, 2017

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]}

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