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Mathematics 18 Online
OpenStudy (anonymous):

how do you find the minimum or maximum value of a quadratic function

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

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

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 = ?

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

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

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