Question 655522: Write each quadratic function in the form f(x) = a (x-h)^2 + k
F(x)= -3x^2 - 12x - 13
Found 2 solutions by nerdybill, DrBeeee:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Write each quadratic function in the form f(x) = a (x-h)^2 + k
F(x)= -3x^2 - 12x - 13
Complete the square:
F(x)= (-3x^2 - 12x) - 13
F(x)= -3(x^2 + 4x) - 13
F(x)= -3(x^2 + 4x + 4) - 13 + 12
F(x)= -3(x+2)(x+2) - 1
F(x)= -3(x+2)^2 - 1
Answer by DrBeeee(684)  (Show Source):
You can put this solution on YOUR website! Puting F(x) into the form of f(x), is called completing the square.
To do this, start with the first two terms of F(x),
(1) -3x^2 - 12x and factor out the coefficient of x^2 as follows,
(2) -3*{x^2 + 4*x}
This can be shown to equate to
(3) -3*{(x+2)^2 - 4}
We can show this by FOILing (x+2)^2 in (3), and get
(4) -3*{x^2 + 4*x + 4 - 4} or
(5) -3*{x^2 + 4*x}, which is the same as (2)
Therefore we can replace the first two terms of F(x) with (3), and get
(6) F(x) = -3*{(x+2)^2 - 4} - 13, which is exactly the same as the given F(x).
Notice that we have created a shifted variable (x+2)^2 to replace the plain x^2 of F(x) as required by the form of f(x), thus it is called completing the square.
Now simplify (6) and get
(7) F(x) = -3*(x+2)^2 -3*(-4) - 13 or
(8) F(x) = -3*(x+2)^2 - 1
Note that (8) matches the form given by
(9) f(x) = a*(x-h)^2 + k
where
a = -3,
h = -2, and
k = -1
Comment: The neat part about this form of the quadratic is that the point (h,k) is the turning point of the parabola. And x = h is the axis of symmetry!
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