Question

I need to take half the coefficient term of the x term and square the result. f(x)=1/3x^2+x+1

Answer 1

\[f(x)=\frac{ 1 }{ 3 }x^2+x+1\]

Answer 2

Simple, but I feel like I'm doing it wrong. Please explain.

Answer 3

@rajathsbhat @kery

Answer 4

ultimately what i need to do is write the given expression in form f(x)=a(x-h)^2+k, but i believe this is the first step

Answer 5

^That clears things up.

Answer 6

yes its about identifying the vertex and all the examples are shown without fractions

Answer 7

Ok first write F(x) as \[\frac{1}{3}(x^{2}+3x+3)\]

Answer 8

Now does it look familiar?

Answer 9

i dont understand where is half the coefficient, then its square

Answer 10

thats what it says to do first

Answer 11

and then to add and subtract that value

Answer 12

i dont want to get to far ahead but after that it says to factor the first three terms aas perfect square trinomial, those are the steps in order

Answer 13

Yes. keep that 1/3 outside & do everything you just said to \((x^{2}+3x+3)\)

Answer 14

i just dont understand how you got that

Answer 15

would be final but i just cant get it \[f(x)=\frac{ 1 }{3}(x+\frac{ 3 }{ 2 })^2+\frac{ 1 }{4 }\]

Answer 16

\[\large \begin{align}F(x)&=\frac{1}{3}(x^{2}+3x+3)\\&=\frac{1}{3}[x^{2}+3x+3+(\frac{3}{2})^{2}-(\frac{3}{2})^{2}]\\&=\frac{1}{3}[(x^{2}+3x+(\frac{3}{2})^{2})+3-(\frac{3}{2})^{2}]\\&=\frac{1}{3}[(x+\frac{3}{2})^{2}+\frac{3}{4}]\\&=\frac{1}{3}(x+\frac{3}{2})^{2}+\frac{1}{3}(\frac{3}{4})\\&=\frac{ 1 }{3}(x+\frac{ 3 }{ 2 })^2+\frac{ 1 }{4 }\end{align}\]

Answer 17

Do you see what i did?

Answer 18

so \[\frac{ 3 }{ 2 }\] is the result of half of the coefficient, and then squared?

Answer 19

i thought half of 1/3 would be 1/6 and the square of 1/6 would be 1/36 - what am i doing wrong there

Answer 20

1/3 is not the co-efficient of x for you to do that.

Answer 21

so what is? i thought coefficient was value associated with x

Answer 22

the way you should do these problems is first separate out the co-efficient of x^2. You see, X^2 should not have anything as its co-efficient. That screen-clip is a simple example that's all.

Answer 23

is there anyway you can give a word description for those steps you did? simple terms would be ok i just have trouble recognizing what you did from step to step

Answer 24

sure. I'll do that.

Answer 25

1. I take 1/3 common from every term & put it outside the whole equation. Do you get it?

Answer 26

Yes i think i am getting it thank you for your help. I will practice another question or two.

Answer 27

yes. you do that. Good luck :)