anonymous
  • 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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
what does "one is 7" mean?
anonymous
  • anonymous
first coordinate of the vertex is always \(-\frac{b}{2a}\) in your case \(a=-\frac{1}{2},b=7\)
anonymous
  • anonymous
ooh i see "one is 7" means the first coordinate is 7

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anonymous
  • anonymous
\[f(7)=-\frac{1}{2}\times 7^2+7\times 7-2\]
anonymous
  • anonymous
that was supposed to be x=7 not one is 7
anonymous
  • anonymous
i see exactly what you are missing you did not subtract 2 at the end
anonymous
  • anonymous
if I subtract 2 from \[\frac{ 49 }{ 2}\] how does that equal \[\frac{ 45 }{ 2 }\]?
anonymous
  • anonymous
?
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
  • anonymous
ok lets go slow
anonymous
  • 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
  • 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
  • anonymous
you lost me entirely
anonymous
  • 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
  • anonymous
typo there , the second line should have 2 49's in it
campbell_st
  • campbell_st
just remember \[-2 =- \frac{4}{2}\]
campbell_st
  • campbell_st
so you have \[\frac{49}{2} - \frac{4}{2} = \frac{45}{2}\]
anonymous
  • 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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