what is the equation of the parabola in vertex form? x^2+6x-y+3=0
@mayankdevnani
vertex form is \[y=a(x-h)^2+k\] where the vertex is (h,k) you can complete the square for this
So put it into standard form first, solve for y
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)
Complete the square, then you can do some algebra and put it into that form
im not exactly sure how to do that
Ok, so as I mentioned before first put it into standard form, solve for y, what do you get
\[x^2+6x-y+3=0\]
What happens if you add y on both sides?
then you get x^2+6x+3=0+y
Right, so we have solved for y and now it's in standard form \[y=x^2+6x+3\]
now we need vertex form
Right, we do this by completing the square, since the coefficient for x is 1 we can do it.
which means...?
Completing the square means we take the standard form and transform it into what I had shown above.
Which is the vertex form \[y=a(x-h)^2+k\]
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
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
9
Right, so now we put it into the brackets but to balance it out we must subtract it by 3, so we have \[y=(x^2+6x+ \color{red}{9})+3 -\color{red}{9}\]
Now you can factor what's inside the brackets
Do you know how to factor?
not quite sure
Essentially you have to find two numbers that add up to 6 and multiply to give 9
_+_=6 _x_=9 What are they
3
Right, since both are 3 we have \[y=(x+3)(x+3)-6 \implies y = (x+3)^2-6\]
Now it's in vertex form, I think you can handle it from here :)
so its D?
yeah
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