Question 1130505: For each of the following curves, find the values of x for which y<0.
a) y=2x^2 +4x +1
b) y=-2x^2+5x+3
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 2x^2+4x+1<0
x=(1/4)(-4+/- sqrt (16-8) or (-1+/-.5 sqrt(2))
Numerically, this is between -1.707 and -0.293 or (-1-.5 sqrt (2), -1 +.5 sqrt(2)) ANSWER
-2x^2+5x+3=0 multiplying by -1, 2x^2-5x-3=0
factors into (2x+1)(x-3)=0, so x=-1/2 and x=3 are critical points.
x=-1 testing and y=-4, so to the left of x=-1/2 , y <0
in between use 0 and the function equals 3, so that part is greater than 0.
above 3, use 4 and the function equals -9, so that part is less than 0. Interval (-oo, -1/2) U (3, oo)
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