Any help is appreciated. I'm very confused and I don't know where to find help!
2007-09-09 16:38:56 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You complete the square. To do so, first consider the equation of y=x^2-4x+1 as y = x^2 -bx + c
Now you add and subtract (b^2/4), or +4 and -4. This gives:
y = (x^2 -4x + 4) + (1-4)
The first parantheses can be factored into: (x-2)^2
So k = 1-4 = -3
So the form is:
y = (x-2)^2 -3
2007-09-09 16:48:30 · answer #1 · answered by J Z 4 · 0⤊ 0⤋
The first form is called the standard form for a parabola, the second is the vertex form.
You complete the square this way
1) y-1 = x^2 - 4x, leave x's on one side
2) y-1 = x^2 - 4x + WHAT, so that the right hand side is a perfect square? Hint: take 4, the coefficient of the x, divide by 2, (=2), then square it, so 4. Add to both sides or
y-1 +4 = x^2 - 4x + 4. Note, you did not change anything, since you added to both sides. Simplify
3) y + 3 = (x-2)^2. Check that the right side is really a perfect square.
4) Write as y - -3 = (x-2)^2. WHY?
5) the Vertex of the parabola is (2,-3). What is h and k in your formula? The vertex is (h,k).
2007-09-09 23:46:23 · answer #2 · answered by pbb1001 5 · 0⤊ 0⤋
You should be. This is a operation called completing the square, and ANY parabola of the form ax^2+bx+c can be solved for roots using this form. If y=0, then -k = (x-h)^2 and (x-h)= +/- sqrt(-k). Finally x = h +/- sqrt(-k).
Given this function, if we had x^2-4x+4, this would be a perfect square (x-2)^2. We can convert what we have to this.....
y= x^2-4x+4-3.
Then y= (x-2)^2 - 3. So here, h=2 and k=-3
2007-09-09 23:49:19 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
Well, we must get a (x-h)^2 term in there somewhere, so start with x^2-4x. Complete the square to get x^2-4x+4 = (x-2)^2. But we added 4 to our function, so we must subtract 4 from the 1 in x^2-4x+1 to get -3. If you don't subtract 4 from the 1, your function has changed. Anyhow, we have
y=x^2-4x+1 = (x-2)^2-3.
So clearly, h=2 and k=-3.
2007-09-09 23:47:09 · answer #4 · answered by brianhawking25 1 · 0⤊ 0⤋
I call the first form standard and the second one vertex form. To get it, complete the square in the first function:
y = (x^2-4x+4) +1 -4
y = (x-2)^2-3
Vertex at (2, -3)
2007-09-09 23:49:03 · answer #5 · answered by Ira R 3 · 0⤊ 0⤋
y=(x-h)^2+k
y=x^2-2xh+h^2+k
from this, we can see that
2h = 4 and h^2+k = 1
h=2 and k=1-4=-3
Thus ans is y=(x-2)^2-3
2007-09-09 23:47:15 · answer #6 · answered by epkw 2 · 0⤊ 0⤋
h = the coefficient of x divided by twice the coefficient of x^2 every time. You should be able to calculate k from there.
2007-09-09 23:46:24 · answer #7 · answered by Tom K 6 · 0⤊ 0⤋