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Mathematics 14 Online
OpenStudy (kittykr98):

what is the equation of the parabola in vertex form? x^2+6x-y+3=0

OpenStudy (kittykr98):

@mayankdevnani

OpenStudy (astrophysics):

vertex form is \[y=a(x-h)^2+k\] where the vertex is (h,k) you can complete the square for this

OpenStudy (astrophysics):

So put it into standard form first, solve for y

OpenStudy (kittykr98):

it may help if i list the answer choices.... A... (x+6)=(y+6)^2 B... (x+36)=(y+33)^2 C... (x+9)^2=(y-3)^2 D... (x+3)^2=(y+6)

OpenStudy (astrophysics):

Complete the square, then you can do some algebra and put it into that form

OpenStudy (kittykr98):

im not exactly sure how to do that

OpenStudy (astrophysics):

Ok, so as I mentioned before first put it into standard form, solve for y, what do you get

OpenStudy (astrophysics):

\[x^2+6x-y+3=0\]

OpenStudy (astrophysics):

What happens if you add y on both sides?

OpenStudy (kittykr98):

then you get x^2+6x+3=0+y

OpenStudy (astrophysics):

Right, so we have solved for y and now it's in standard form \[y=x^2+6x+3\]

OpenStudy (kittykr98):

now we need vertex form

OpenStudy (astrophysics):

Right, we do this by completing the square, since the coefficient for x is 1 we can do it.

OpenStudy (kittykr98):

which means...?

OpenStudy (astrophysics):

Completing the square means we take the standard form and transform it into what I had shown above.

OpenStudy (astrophysics):

Which is the vertex form \[y=a(x-h)^2+k\]

OpenStudy (astrophysics):

So to start off, we move the +3 outside and put brackets around \[x^2+6x\] as follow \[y=(x^2+6x~~~~~)+3\] notice I have put space between the 6x and bracket

OpenStudy (astrophysics):

The number that goes into the space what we do is take 1/2 of the 6 and then square it, which is the middle term (6x). \[\left( \frac{ 6 }{ 2 } \right)^2 = ?\] what does that give us

OpenStudy (kittykr98):

9

OpenStudy (astrophysics):

Right, so now we put it into the brackets but to balance it out we must subtract it by 3, so we have \[y=(x^2+6x+ \color{red}{9})+3 -\color{red}{9}\]

OpenStudy (astrophysics):

Now you can factor what's inside the brackets

OpenStudy (astrophysics):

Do you know how to factor?

OpenStudy (kittykr98):

not quite sure

OpenStudy (astrophysics):

Essentially you have to find two numbers that add up to 6 and multiply to give 9

OpenStudy (astrophysics):

_+_=6 _x_=9 What are they

OpenStudy (kittykr98):

3

OpenStudy (astrophysics):

Right, since both are 3 we have \[y=(x+3)(x+3)-6 \implies y = (x+3)^2-6\]

OpenStudy (astrophysics):

Now it's in vertex form, I think you can handle it from here :)

OpenStudy (kittykr98):

so its D?

OpenStudy (astrophysics):

yeah

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