can someone plz explain and help me put this into standard form >>> Y=0.05x^2+x+1?
Y=0.05x^2+x+1 This is in standard form for a parabola, which you have. Do you mean vertex form? vertex form = y= a(x-h)2+k
yes thank you now I know why no one answers my question srry.
Ok one sec let me type it out. You just want the answer or would you want me to explain?
explain. and show.
kk give me a sec
okay.
is the vertex (.25,0)?
Grrr was typing out a huge tutorial and some how it got wiped lol
Got to the x of the vertex, which will be 10
omg. that sucks.srry
so my answer is wrong?
Ok we need to find the vertex. To find the vertex we use the Y=0.05x^2+x+1 equation and the formula \( \frac{b}{2(a)} \) b = x =1 a = .05 \[ \frac{1}{2(.05)} \] \[ \frac{1}{.1} = 10 \] Ok our x for our vertex = 10 now to find the y for our vertex point we need to insert 10 for x in the equation Y=0.05x^2+x+1 \[ Y=0.05(10)^2+10+1 \] \[ Y=0.05(100)+10+1 \] \[ Y=5+10+1 \] \[ Y=11 \] Now we have our y = 11 so our vertex is (10,11) so to put this in vertex form we plug our points in to y= a(x-h)^2+k a = 0.05 x = x h = vertex x 10 k = vertex y 11 Now just plugin \[y= a(x-h)^2+k \] \[y= 0.05(x-10)^2+11 \]
just give me a sec to read it over plz
Sorry, where ever you see 11 it should be 16
I was rushing that time around lol
oh okay so y=16 then?
yes
lol that's fine , I'm glad you took the time to help me out.
so the axis is -16?
Also, remember the x of the vertex can be found using \( \frac{b}{2(a)} \) which will give you the x of the vertex
no wait the axis would be -10.
is that right?
Yes
thank god.
okay , well thanks so much for helping me.
No problem anytime.
I am here to help with what I can
I know math can be frustrating lol
yah it really is , I have a short fuse for this stuff sometimes. lol
It is a lot harder too and more frustrating when you don't have anyone around to ask questions. That is why I like places like this :-)
ikr , I totally agree, I'm glad I found this place.
Ok have a good one. I will see ya around
kk. u2.
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