Mathematics
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OpenStudy (anonymous):
how do you find the minimum or maximum value of a quadratic function
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hartnn (hartnn):
calculus ?
OpenStudy (anonymous):
Find the vertex
hartnn (hartnn):
or without calculus ?
OpenStudy (anonymous):
well ill let @harttnn handle it :D
OpenStudy (anonymous):
algebra
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hartnn (hartnn):
so say u have a quadratic function like f(x)=ax^2+bx+c
first, u need to bring it in the form of f(x)= a(x-h)^2+k
then max. or min occurs at vertex h,k
hartnn (hartnn):
any particular question ?
OpenStudy (anonymous):
2x^2+20x-9
hartnn (hartnn):
or simply u can use that max. or min of f(x)=ax^2+bx+c occurs at -b/2a
OpenStudy (anonymous):
thats what i was thinking
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OpenStudy (anonymous):
but it doesnt give me the result
hartnn (hartnn):
so, Compare your quadratic equation with \(f(x)=ax^2+bx+c\)
find a,b,c.
what result u got ?
OpenStudy (anonymous):
-59
OpenStudy (anonymous):
a=2 b=20 c=-9
hartnn (hartnn):
so -20/4 = ?
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OpenStudy (anonymous):
-5
OpenStudy (anonymous):
and then you plug it in?
OpenStudy (anonymous):
hahaha wow!
hartnn (hartnn):
yes, thats correct.
OpenStudy (anonymous):
i have another question
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hartnn (hartnn):
ask..
OpenStudy (anonymous):
to graph f(x)=4x^2-8x+1 starting with the graph y=x^2
hartnn (hartnn):
so u have graph of y=x^2
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
i have to use transformation in order to graph the new function