Question

find the vertex of the parabola by completing the square x^2-6x+8=y
the choose are
A.(-3,-6)
b(3,-1)
C(-3,1)
D(-1,3)

Answer 1

Do you know how to complete a square?

Answer 2

no

Answer 3

Completing the square is a method used with quadratics, whose graphs are parabolas. So may be used in either case.
The end result of the completing the square techniques should be a quadratic equations/expression in vertex form:
$$y=k(x-p)^2+q$$
where (p,q) is the vertex of the parabola, the vertical line x=p is the axis of symmetry, and the sign of k determines whether the parabola opens up or down.

The method completes an ``incomplete" square:
$$ x^2 \pm ax$$
by realizing that
$$ \left(x\pm \frac{a}{2}\right)^2=x^2\pm ax + \left(\frac{a}{2}\right)^2$$
for example
$$ \left(x+ 5\right)^2=x^2 +5x+5x + \left(5\right)^2= x^2 + 10x + 25$$
and therefore
$$ x^2 + 10x + 25= \left(x+ 5\right)^2$$
In this case we have a ``complete square", but what if we had an incomplete square?
Here is an incomplete square. Notice that we could complete it by adding 25 to it.
$$ x^2 + 10x=f(x) $$
However, the rules of operations do NOT allow us just to add numbers to one side of an equation. It DOES allow us to add zero as much as we want though. In this case, +25 -25 is the ``zero" that we want to add.
$$ x^2+10x + 25 -25=f(x)$$
Notice we HAVE NOT changed the equation at all. We could simplify the +25 and -25 to get back to our original equation. Please do not though, because with the positive part of the ``zero" that we have added we can complete in incomplete square:
$$(x-5)^2 -25 =f(x)$$
and convert the equation to the vertex form. We know that this parabola is open up, has a vertex at (5,-25), and an axis of symmetry of x=5. If we wanted to know where the `zeros' or `roots' of the parabola was, we set f(x) to zero and solve.
$$(x-5)^2 -25 = 0 \rightarrow\text{Add 25 to both sides}$$ $$(x-5)^2 = 25 \rightarrow\text{sqr. root both sides}$$ $$ x-5 = \pm \sqrt{25}$$ $$ x-5 = \pm 5 \rightarrow\text{Add 5 to both sides}$$ $$ x = +5\pm 5 = 10 \text{ and } 0 $$

Answer 4

huh

Answer 5

That was just for reference, I wrote that a while ago when someone else didn't know what "completing the square" was.

Answer 6

oh ok

Answer 7

The incomplete square given to you is
\[x^2-6x\]
The idea is that there is some
\[(x-a)^2\] that gives you $$x^2-6x \text{ and something} $$
For example
\[(x-2)^2 = x^2 -4x +4\]
Can you guess what value for "a" would give you
\[(x-a)^2 = x^2-6x? \text{ and something} \]

Answer 8

so how would i sovle the problem

Answer 9

You need to put the parabola into vertex form
\[y=k(x-p)^2 + q\] because by definition the vertex in this form is (p,q)
You do this by completing the square.
A parallel problem,
If I was given
\[ x^2 -4x +10 =y\] I would
say that \[(x-2)^2 = x^2-4x +4\]
Then rewrite what was given to you as:
\[ x^2 -4x +4 -4+10 =y\]
\[ \bigg(x^2 -4x +4\bigg) -4+10 =y\]
complete the square in the parentheses

\[ \bigg(x-2\bigg)^2 +6 =y\]
and then since this is in Vertex form I know that p=2 and q =6
so the vertex is (2,6)

Answer 10

this is no 2,6 on the chooses

Answer 11

The moderators are very clear that helpers are not allowed to do problems for students. I gave you essentially the same problem with slightly different numbers. You need to look at the problem I did, see how to do it with your problem and do it. It is the closest I can get to doing it for you.

Answer 12

not asking for you to do it for me just asking for help

Answer 13

Ok,
What is the square that you are asked to complete?

Answer 14

x^2-6x+8=y

Answer 15

No, that's the whole equation. The square they want you to complete is
\[x^2 -6x\] which is a piece to the equation. When you complete it you will have
\[\bigg(x^2-6x +n\bigg)-n\] that you would then substitute into the equation.
You need to figure out what value of 'a' will give you this kind of answer if
\[(x-a)^2 = x^2-2a +a^2\]