A quadratic function with a minimum of 5 has zeros at -4 and 2, find the equation of this function.
This is impossible, correct?
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Yes, seems impossible. Most probably they meant $-5$ for the minimum. – Berci May 06 '14 at 23:26
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Or they meant 5 is the MAXimum... – colormegone May 06 '14 at 23:48
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It's impossible. It may not have real zeroes if the minimum is above zero.
If they meant y=-5 as the minimum, then you could use a matrix to solve for $f(x)$
$y=\frac59x^2+\frac{10}9x+\frac{40}9$
If they meant y=5 is a maximum, then:
$y=\frac{-5}9x^2-\frac{10}9x+\frac{40}9$
ayane_m
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