anonymous
  • anonymous
Can someone help me to find a vertex of parabola??? y=4x^2 + 4x -5
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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xapproachesinfinity
  • xapproachesinfinity
\[\text{the technique to find to vertex of the probola given it's standard equation }\\ \text{y=ax^2+bx+c is called completing the square}\]
anonymous
  • anonymous
not the way i do it
xapproachesinfinity
  • xapproachesinfinity
so you need to complete the square for y=4x^2+4x-5

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More answers

xapproachesinfinity
  • xapproachesinfinity
@satellite73 you mean u use calculus?
xapproachesinfinity
  • xapproachesinfinity
@Savannah_Lynn13 hey there interested to do this?
misty1212
  • misty1212
HI!!
Nnesha
  • Nnesha
no he use `-b/2a` formula :D
misty1212
  • misty1212
first coordinate of the vertex is always \(-\frac{b}{2a}\) and if you know the first coordinate you find the second by substitution
Nnesha
  • Nnesha
no he use `-b/2a` formula :D to find x-coordinate of the vertex and then substitute all x's for its value into the equation to find y-coordinate :P
misty1212
  • misty1212
here we have \(a=4,b=4\) put in in to \[\huge-\frac{b}{2a}\] to get the \(x\) value of the vertex
xapproachesinfinity
  • xapproachesinfinity
well that's if he he knows that (-b/2a, f(-b/2a) ) is the vertex
anonymous
  • anonymous
ok so the thing i plug in would be -1/2, correct?
misty1212
  • misty1212
yes
xapproachesinfinity
  • xapproachesinfinity
yes x=-1/2 you need f(-1/2) too
xapproachesinfinity
  • xapproachesinfinity
i prefer the other way which show where thing come from it is okay to use x=-b/2a as long as you know where it comes from
xapproachesinfinity
  • xapproachesinfinity
i though @satellite73 meant to take first derivative lol when he not the way i do it
ShadowLegendX
  • ShadowLegendX
I think I skipped class that day
xapproachesinfinity
  • xapproachesinfinity
hahah lol
xapproachesinfinity
  • xapproachesinfinity
still need help i guess you know thee answer
ShadowLegendX
  • ShadowLegendX
oh completing the square
anonymous
  • anonymous
so i get to\[y=4(-\frac{ 1 }{2})^2+4(-\frac{ 1 }{ 2 })-5\] and no i don't know the anwer
anonymous
  • anonymous
it is arithmetic from here on in
anonymous
  • anonymous
what is \(\left(-\frac{1}{2}\right)^2\)?
anonymous
  • anonymous
yeah and i have a very difficult time with math, so for some reason this is really confusing me
anonymous
  • anonymous
if you don't know, i will tell you, but it is easy to square a fraction, square the top and the bottom
anonymous
  • anonymous
-1/4
anonymous
  • anonymous
no, when you square the minus sign goes away
anonymous
  • anonymous
\[\left(-\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)\times\left(-\frac{1}{2}\right)=\frac{1}{4}\]
anonymous
  • anonymous
now we are at \[y=4(\frac{ 1 }{4})+4(-\frac{ 1 }{ 2 })-5\]
anonymous
  • anonymous
you good from there?
anonymous
  • anonymous
is it -6?
anonymous
  • anonymous
Is this correct?
lochana
  • lochana
http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html
lochana
  • lochana
y=4x^2 + 4x -5 y = 4(x^2 + x) -5 y = 4(x+1/2)^2 -4(1/4) -5 y = 4(x + 1/2)^2 - 6 vertex point = (-1/2, -6)
anonymous
  • anonymous
Thank you :)
lochana
  • lochana
you need to convert the equation into vertex form as mentioned in that web page. hope it helps

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