Answer: it doesnt need an explanation, it just does.
Step-by-step explanation: im not stoopid
Answer:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
Step-by-step explanation:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)
Consider x^2-2/3x+1/3
Remember that (a-b)^2 =a^2-2ab+b^2
x^2=a^2
a=x
-2/3x= -2*x*b
b=1/3
S0 (x-1/3)^2= x^2-2/3x+1/9
x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
In order to do this we need to make them both fractions, or both decimals. It's much easier to make them both decimals, as you can get

into decimal form by simply dividing 1 by 5.

If 1/5 = .20, then 1/5 is greater than .15!
Problem 9
The instructions aren't stated anywhere, but I'm assuming your teacher wants you to find the equation of each parabola.
The vertex here is (h,k) = (-6,-4) which you have correctly determined.
This means

Next we plug in one of the other points on the parabola. We cannot pick the vertex again. Let's pick the point (-4,0) which is one of the x intercepts. We'll do this to solve for 'a'

This means

represents the equation of the parabola in vertex form.
<h3>Answer:

</h3>
========================================================
Problem 10
The vertex is (h,k) = (-2,-6)
So,

Now plug in another point on the parabola like (-1,-8) and solve for 'a'

<h3>Answer:

</h3>
For each equation, you could optionally expand things out to get it into y = ax^2+bx+c form, but I think it's fine to leave it as vertex form.
Replace X in the equation with 0:
f(0) = 3-2(0)
f(0) = 3-0
f(0) = 3