Quadratic Functions, and Completing the Square

[H.M. Murdock]

I had a hard time completing the square of some problems. This is what I did with them, I would appreciate if someone could show me the mistake that I am doing and the right process in order to solve these problems.
Also, here there is an article about completing the square for anyone that dosent remeber about it.
http://www.uncwil.edu/courses/mat111hb/Izs/asolve/asolve.html#compsqr


First problem
Complete the square on y = 3xsquare - x + 1 to write it in the form y = a (x - h)square +k

y = 3xsquare - x +1

3xsquare/3 and - x/3 =


3 (xsquare - 1/3x) +1


(1/3)square = 1/9


3 (xsquare -1/3x + 1/9) + 1 - 1/3


The answer is y = 3 (x + 1/3) square 2/3


But the answer on my book is y = 3 (x - 1/6) square + 11/12
I d like to know what was wrong there.



Second problem
Complete the square on y = xsquare - 6x + 10 to write it in the form y = a (x -h)square +k


y = xsquare - 6x + 10


(xsquare -6x) + 10


(-6/2)square = 9


(xsquare - 6x + 9) 10 - 9


The answer is (x + 3)square + 1


However on my book the aswer is y = (x - 3)square + 1
I don't really know where did that "-3" came from.


Third problem
-What are the x- and y- intercepts for f(x) = 2xsquare + x - 6 ?

This quadratic funtions is in the form " axsquare + bx + c " so I used the formula " h = -b/2a "
in order to solve it.

a = 2 and b = 1 on the function


so h = -1/2(2) = -1/4

k = 2(-1/4) square + (-1/4) - 6

k = 1/8 - 1/4 - 6

k= 49/8

I would appreciate if anyone is able to tell me how to solve these 3 problems.
Thanks a lot in advance.

 

[happyg1]

Hi,
on problem 1, You have to divide ALL of the terms by 3. I did this, and i didn't get what your answer is:
$$\frac{3x^2}{3}-\frac{x}{3}+\frac{1}{3}=0$$
$$x^2-\frac{1}{3}x=-\frac{1}{3}$$
that third term is then:
$$\left(\frac{1}{2}\cdot\frac{1}{3}\right)^2=\frac{1}{36}$$
then add that to both sides:
$$x^2-\frac{1}{3}x+\frac{1}{36}=-\frac{1}{3}+\frac{1}{36}$$
see what you get from there.

Second problem:
you got
$$x^2-6x+9=-1$$
when you factor that you have:
$$(x-3)^2+1$$ (after you add the one back over)

Third problem:
You found the x-value for the minimum, not the x and y intercepts.
For x intercepts, set it equal to zero and solve.
Your y-int. is f(0).
CC

 

[Redbelly98]

First problem:

as happyg1 said, you don't square (1/3). First you divide by 2, and then square that:

[(1/2)*(1/3)]^2 = (1/6)^2 = 1/36

and so

x^2 - x/3 + 1/36
is a perfect square, equal to
(x - 1/6)^2

 

[H.M. Murdock]

First problem:

as happyg1 said, you don't square (1/3). First you divide by 2, and then square that:

[(1/2)*(1/3)]^2 = (1/6)^2 = 1/36

and so

x^2 - x/3 + 1/36
is a perfect square, equal to
(x - 1/6)^2

Hi thanks a lot guys, I appreciate it , but where does the 2 of the division come from?

 

[Dick]

(x+a)^2=x^2+2xa+a^2. So if you see x^2+Ax, if it's part of the complete square (x+a)^2, A=2a. Just equate the coefficients of the linear factors.