## turning points and nature of turning points

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### turning points and nature of turning points

hi,

i need to know how to work this out:

find the turning points and nature of turning points for:
y=(3x-1)^2-4

for this i have expanded the brackets and got 9x^2-6x+1
then i minus 4 to get 9x^2-6x-3=0

however now im not sure what to do from here and this is where i require some help please.
could you explain what i have to do next so i understand it and show me step by step please?

thanks.
harry buckle
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### Re: turning points and nature of turning points

harry buckle wrote:hi,

find the turning points and nature of turning points for:
y=(3x-1)^2-4

for this i have expanded the brackets and got 9x^2-6x+1
then i minus 4 to get 9x^2-6x-3=0

What is a "turning point"? That is, what do YOU think a turning point is? Also, why did you expand the parentheses and subtract 4?

mrtwhs
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### Re: turning points and nature of turning points

mrtwhs wrote:
harry buckle wrote:hi,

find the turning points and nature of turning points for:
y=(3x-1)^2-4

for this i have expanded the brackets and got 9x^2-6x+1
then i minus 4 to get 9x^2-6x-3=0

What is a "turning point"? That is, what do YOU think a turning point is? Also, why did you expand the parentheses and subtract 4?

its the maximum and minimum points on a curve. i did box method to work out (3x-1)^2, wouldnt i do that??
harry buckle
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Joined: Wed Jan 25, 2012 10:53 am

### Re: turning points and nature of turning points

Harry, do you realise that 9x^2 - 6x - 3 = 0 is same as 3x^2 - 2x - 1 = 0 ?
Can you factor that?
I'm just an imagination of your figment...
Denis
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### Re: turning points and nature of turning points

Only the valid values for x and the x - coordinate of the top are the same there.
(3x-1)^2 - 4 has a minimum for 3x - 1 = 0 and is equal to
(3x-1)^2 - 2^2 = (3x - 1 - 2)(3x - 1 + 2) = 3(x - 1)(3x + 1)
Hoempa
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### Re: turning points and nature of turning points

would x = 1 and x= -0.3?
harry buckle
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### Re: turning points and nature of turning points

Are you expected to use differential calculus?

I ask because normally finding relative extrema is done in the first semester of calculus, but a parabolic function in vertex form has its turning point given to you.

You were given:

$y=(3x-1)^2-4$

We merely need to factor out the coefficient of x, which is being squared, to write:

$y=9$$x-\frac{1}{3}$$^2-4$

A parabola in the form:

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

has its turning point at (h,k) and this turning point will be a minimum if a > 0 and a maximum if a < 0.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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Location: St. Augustine, FL., U.S.A.'s oldest city

It's a bit odd to ask for the turning points (plural) when there is only one. There was no reason to add "= 0".

By first considering the corresponding problem for y = (3x - 1)², it becomes "obvious" that (3x - 1)² - 4 has a minimum at (1/3, -4) and no maximum.

Since y = (3x - 1)² - 2² = (3x - 1 - 2)(3x - 1 + 2), the x-axis intercepts are at x = -1/3 and x = 1. By symmetry, the x-coordinate of the turning point is the mean of these values, which is 1/3. The corresponding y-coordinate is -4.
skipjack
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