Best answer to whoever first tell me what the turning point equation is! :)?

Question

maths.

Answer 1

For a quadratic function
f(x) = ax² + bx + c,

the turning point is (-b/2a, f(-b/2a))
and
the turning point equation is
f(x) = a(x + b/2a)² + f(-b/2a)

You can also get the quadratic equation into the turning point form by completing the square in x.
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EXAMPLE.

f(x) = 2x² + 4x + 1

-b/2a = -4/4 = -1
f(-b/2a) = f(-1) = 2(-1)² + 4(-1) + 1 = 2 - 4 + 1 = -1

=> The turning point is (-1, -1) and the turning point equation is f(x) = 2(x+1)² - 1
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Let's also find the turning point equation by completing the square.

f(x) = 2x² + 4x + 1

We are only interested in the terms involving x. Factor 2 from these terms (so that x² will be on its own - it will make things easier):

f(x) = 2[x² + 2x] + 1

Now look at the thing in the square brackets. It looks ALMOST like x² + 2x + 1 (which is (x+1)²).
We ADD + 1 to get x² + 2x + 1, but then need to subtract -1, to keep the equation in balance:

f(x) = 2[x² + 2x + 1 - 1] + 1
f(x) = 2[(x+1)² - 1] + 1

now break the square brackets

f(x) = 2(x+1)² - 2 + 1

f(x) = 2(x+1)² - 1

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Hope this helps.

Answer 2

Consider:-
f(x) = 3x² + 4
f `(x) = 6x = 0 for turning points
6x = 0
x = 0
f(0) = 4
(0 ,4) is a turning point.
f "(0) = 6 is + ve thus:-
(0 , 4) is a MINIMUM turning point.
This is a parabola.
Hope this example helps.