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April 20, 2014

April 20, 2014

Posted by **Haylee** on Friday, May 18, 2007 at 9:57am.

For Further Reading

algebra - Reiny, Thursday, May 17, 2007 at 10:34pm

I am sure neither your text nor your teacher asked you to "solve the polynomial y=2x^2-4x+1 "

are you graphing the function, or are you solving the quadratic y=2x^2-4x+1 ?

I am trying to

solve the quadratic sorry.

2x^2-4x+1 = 0 does not factor, so use the formula

x =(4 ± sqrt(16-4(2)(1)))/4

=(4 ± √8)/4

=(2 ± √2)/2

if you are looking for the vertex of

y=2x^2-4x+1

a quick way is to find the x of the vertex by x= -b/(2a) = 4/4 = 1

now sub x=1 back in your function to get the y of the vertex.

the other way would be to complete the square for 2x^2-4x+1

To make sure i did this right does y=1 also?

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