xjgrox said:
Yes I got a parabola as well, and Alright I'll take another look and see. The main thing I am getting stuck on is completing the square, my teacher gave me the answer without me fully understanding how to get it. He told me it was f(x)=4(x+(1/4))2-(1/4) ...that being the completed square
Think of k as being any number, then
(x+k)^2=x^2+2kx+k^2
But to keep it simple, let's ignore the details of this result and just focus on what's most important. It is of the form x
2 + (number)x + (number). We of course want to turn a general quadratic such as x
2 + bx + c (b,c are now numbers) into the completed square because that form is easier to work with, mostly because the completed square form involves just a single x in the entire expression.
Now, since (x+k)^2=x^2+2kx+k^2, if we had to complete the square on, say, x^2+3x+1, we'd want to choose a value of k such that x^2+2kx+k^2=x^2+3x+1 which means that we need both of these results to hold: 2k=3, k^2=1, but in the first equation, k=3/2, while in the second k=1 or -1. We have no common solutions between the two equations, so there is no such k that we can use to turn x^2+3x+1 into (x+k)^2.
What if we just choose to satisfy one of the equations though? Let's make the number 1 disappear, so we will choose k=1 or k=-1 (let's choose k=1 first), then
(x+k)^2=(x+1)^2=x^2+2x+1
Therefore,
x^2+3x+1=x^2+2x+1+x=(x+1)^2+x
So now we've removed the constant which in this case is 1, but we still have x left over which doesn't help us with graphing.
If we choose k=-1, then
(x+k)^2=(x-1)^2=x^2-2x+1
Therefore,
x^2+3x+1=x^2-2x+1+5x=(x-1)^2+5x
Which as before, still doesn't help us. So rather than choosing k such that we remove the constant, let's choose k such that we remove the x terms (which is what our goal is anyway). If we choose k=3/2 then
(x+k)^2=(x+3/2)^2=x^2+3x+9/4
Therefore,
x^2+3x+1=x^2+3x+9/4-5/4=(x+3/2)^2-5/4
Which is great! Because now we can graph it more easily since a parabola of the form
y=(x+a)^2+b
Has its vertex at (-a,b). The reason why it's that point is because n
2 is the smallest value it can be when n=0. Any other value of n gives us n
2>0, so when n=0, n
2=0. But in our case, n=x+a, and again, (x+a)
2 is the smallest value it can be when the thing being squared = 0, hence when x+a=0, or x=-a. So at x=-a we have the minimum of the parabola, and when we plug x=-a into the equation we get y=(-a+a)^2+b=0^2+b=b hence y=b is the y value at this x value, hence we have (-a,b) as the vertex of the parabola. We can also easily find where it cuts the x-axis which is when y=0. We get that
(x+a)^2+b=0
(x+a)^2=-b
x+a=\pm\sqrt{-b}
x=-a\pm\sqrt{-b}
So if b is a negative value, then -b is positive and that means it then cuts the x-axis. If b is positive then we can't square root a negative and that means it doesn't cut the x-axis which means the parabola is entirely above the x-axis.
So in summary, how do we complete the square? We choose a value of k in (x+k)^2=x^2+2kx+k^2 such that we delete the x value, which means 2k is our coefficient of x (number multiple of x). We then choose that value of k, expand the completed square (x+k)
2 and then split up the constant such that one part equals our expanded k
2, and the other part is chosen such that it equals our constant. If you're unsure, just follow what I did earlier.
Finally, if your coefficient of x
2 isn't 1, then all you need to do is factor out the coefficient first, and then everything else is the same. For example,
4x^2+2x=4(x^2+x/2)
And then do everything the same in the brackets, and finally at the end you should arrive at the answer 4((x+1/4)^2-1/16) and you then expand again to get 4(x+1/4)^2-1/4
