Question #5508c

1 Answer
Oct 29, 2017

#6(x+3)^2 - 53 = 0#

Note that the parabola's vertex is at the point #(-3, -53)# (you may need to scroll down on the graph to see):

graph{6(x+3)^2 - 53 = y [-10, 10, -5, 5]}

Explanation:

A quadratic is in vertex form when it is something like this:

#y = a(x-h)^2 + k#

It is so named because the point #(h, k)# is the very tip of the parabola, i.e. the vertex.

The easiest way to get into vertex form is to use the method of completing the square. This essentially means you add something to this polynomial to make it a perfect square.

Step 1: Factor Out Leading Coefficient
So we'd have to factor out a 6 here, which would leave us with:

#6(x^2 + 6x -1/6) = 0#

Step 2: What to add?
This is probably the hardest part of this entire process. You need to find something to add to that quadratic above to make it a perfect square. To do this, let's recall the expansion of a perfect square:

#(a + b)^2 = a^2 + 2ab + b^2#

We know that #a# is just #x#. Now, we need to have some #b# such that #2(x)b = 6x#. So, let's solve:

#=> 2xb = 6x#
#b = 3#

Yay! However, we need to add something that makes the quadratic above #a^2 + 2ab + b^2#. In other words, we need to add something that turns that #-1/6# into #b^2#, or #9#. So, we'll need to add:

#9 - (-1/6) = 55/6#.

However, note that this is what we're adding to the quadratic inside the parenthesis. Because everything in there is multiplied by #6#, what we're really adding is #55#.

Step 3: Actually Add
Adding #55/6# inside the parenthesis gives:

#6(x^2 + 6x -1/6 + 55/6) = 0#

But wait, are we just allowed to add numbers to one side of the equation? The answer is no! What we do to one side of the equation, we must do to the other side. Therefore, we'd add #55# to the right hand side of the equation as well:

#6(x^2 + 6x -1/6 + 53/6) = 53#

Step 3: Simplify
Some quick cleanup gives:

#6(x^2 + 6x + 9) = 53#

We know that #x^2 + 6x + 9# is really just an expansion of #(x + 3)#^2, so we can condense that:

#6(x+3)^2 = 53#

Lastly, we just subtract 53 from both sides, to achieve our desired form:

#6(x+3)^2 - 53 = 0#

And that is your final answer!

Useful Video:

Hope that helped :)