Question 143395: Find the equation of the parabola (second degree polynomial) that passes through the points (-3,0), (4,0), and (0,2)
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! First thing to notice is that two of the points are on the x-axis and that tells us that we have a parabola whose axis is perpendicular to the x-axis, i.e. vertical, and that the x-coordinates of these two points are the zeros of the desired polynomial function.
Knowing that we have two zeros for a 2nd degree polynomial, we can derive A quadratic function simply by multiplying  . However, this misses the mark because if you calculate the y-intercept you get (0,-12) instead of the desired (0,2).
Fortunately, you can multiply any polynomial by any constant and not change the zeros.  ,  , and  all have the same zeros, namely 1 and -5.
So for the function in question, we need to answer, "What can we multiply by so that the constant term will be 2?" Answer:  .
So:  is the desired function.
Check:
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