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Question 49123: indicate whether the graphs of the following will have a maximum or a minimum.
1. f(x)=x exponent 2-9
2.f(x)=8x-3x exponent 2
3.f(x)=-(3-x)exponent 2
Answer by Born2TeachMath(20)   (Show Source):
You can put this solution on YOUR website! Your teacher is trying to get you to realize that the coefficient of the x^2 term in a quadratic graph determines whether the graph of a parabola will open up (and have a bottom - or minimum value) or open upside down (and have a top - or maximum value).
If the coefficient of the x^2 term is positive, then the graph of the parabola opens upwards, and will have a minimum value.
If the coefficient of the x^2 term is negative, then the graph of the parabola opens downward, and will have a maximum value.
The up or down direction does not depend on either the sign of the x-term or the constant at the end. These two have other significance which you'll get into later, I'm sure.
So, to answer your questions:
1. f(x) = x^2 - 9 will open up and have a minimum, since the coefficient of the x^2 is +1.
2. f(x) = 8x - 3x^2 will open down and have a maximum, since the coefficient of the x^2 is -3
3. f(x)=-(3-x)^2
This one is a bit trickier. You really should simplify the left side by multiplying it out:
-(3-x)^2 = -(3-x)(3-x) definition of squaring = multiplying by itself
= -(9-6x+x^2) FOIL it out.
=-9+6x-x^2 distribute the (-).
Therefore, the parabola will open down and have a maximum, since the coefficient of the x^2 is -1
Don't try to just guess on #3!
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