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- anonymous

f(x)=-1/2x^2+7x-2. x=7. Im trying to find the vertex, one is 7, and the other is 45/2. I keep getting 49/2. what am I missing?

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- anonymous

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- anonymous

what does "one is 7" mean?

- anonymous

first coordinate of the vertex is always \(-\frac{b}{2a}\)
in your case \(a=-\frac{1}{2},b=7\)

- anonymous

ooh i see "one is 7" means the first coordinate is 7

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- anonymous

\[f(7)=-\frac{1}{2}\times 7^2+7\times 7-2\]

- anonymous

that was supposed to be x=7 not one is 7

- anonymous

i see exactly what you are missing
you did not subtract 2 at the end

- anonymous

if I subtract 2 from \[\frac{ 49 }{ 2}\] how does that equal \[\frac{ 45 }{ 2 }\]?

- anonymous

?

- anonymous

the vertex is (7, 45/2). I cant figure out how its not 49/2. the class im in offers no explanation as to how it gets the answers, so im floored on how 7x7 equals 45.

- anonymous

ok lets go slow

- anonymous

did you get that you find the second coordinate of the vertex by finding \[f(7)=-\frac{1}{2}\times 7^2+7\times 7-2\]

- anonymous

yes, I got that far find and this is what I did, in order (7)^2=49. 7x7=49-2=47 so then I have \[-\frac{ 1 }{ 2 }(49)+47\]

- anonymous

you lost me entirely

- anonymous

\[-\frac{1}{2}\times 7^2+7\times 7-2\\
-\frac{1}{2}\times 49+47-2\\
-\frac{49}{2}+\frac{98}{2}-2\\
\frac{49}{2}-2\\
\frac{45}{2}\]

- anonymous

typo there , the second line should have 2 49's in it

- campbell_st

just remember
\[-2 =- \frac{4}{2}\]

- campbell_st

so you have
\[\frac{49}{2} - \frac{4}{2} = \frac{45}{2}\]

- anonymous

okay thank you I didn't know how to get the second fraction. I see it know. I wish my class program would just explain it seesh. thanks again!

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