Turning point

[Studyinghard]

Heyz

looks like I have completely forgotten how to complete the square and find out the turning point for quadratics.

a) Find out TP for
b) Complete the square of
c)

With question C would it just be 0 and -1/2 and I dont have to complete the square or anything?

Thanks

[Damo17]

Heyz

looks like I have completely forgotten how to complete the square and find out the turning point for quadratics.

a) Find out TP for
b) Complete the square of
c)

With question C would it just be 0 and -1/2 and I dont have to complete the square or anything?

Thanks

a) find the x value of the TP using
then

so TP at

b)

so TP at

c) yes, since translation is only dealing with y-axis with no other transformations to consider, the TP is at

[brightsky]

a) The turning point for any quadratic is

    Hence, the turning point of is

    This equals to

    Substitute into .
 
   

    and

    So minimum turning point is at

    Alternatively, you can complete the square using the turning point form. Such as in b).

b)
   
   

   

   

    Hence, if you want to find the turning point of this, the minimum turning point is at .

c) Yes, your correct for C. It's just a normal parabola moved 1/2 down the y-axis.

[Studyinghard]

Hm..i dont know why but my maths quest textbook gives me a completely different answer for question A.


a)

[brightsky]

Don't think 1729/144 is right... 144 shouldn't be a denominator anyway you look at it.

[Stroodle]

Yeah, your text must be wrong. Brightsky's answer is correct.

[EvangelionZeta]

Funny how after a whole year of Methods 3/4, I don't remember using the -b/2a thing AT ALL.  xD

Seconding that brightsky's answer for a is correct though.

[kyzoo]

^ me neither lol, i just used {x:dy/dx = 0}

[Studyinghard]

i know right. i mean i just finished 1/2 and now doing 3/4 which starts of with revising units 1/2 im just going completely blank on this crap.

[n.f]

^ me neither lol, i just used {x:dy/dx = 0}

Has it ever come up in exams to find turning points via the other method? Or just find them through a method (ie. x:dy/dx = 0)

[kyzoo]

I did 30 practice exams and I don't remember using a method other than {x:dy/dx = 0} to find turning points.

[EvangelionZeta]

I did 30 practice exams and I don't remember using a method other than {x:dy/dx = 0} to find turning points.

This applies to me as well.  :p

[GerrySly]

^ me neither lol, i just used {x:dy/dx = 0}

Has it ever come up in exams to find turning points via the other method? Or just find them through a method (ie. x:dy/dx = 0)

It's just find them through a method. Just remember and and then after doing a lot of them you can just pick your preference (the one you're quickest at)

[kamil9876]

it comes easily if u know parabola is symmetric. y=ax^2+bx+c cuts through the line y=c at:

0=x(ax+b)

x=0 and at x=-b/a

thus the turning point is at x=-b/2a.

===================================================

Likewise the turning point of y=(x-a)(x-b) is (a+b)/2. (again the midpoint of two points with equal y-values(in this case both with y=0)). Again, that's how the -b/2a appears in the quadratic formula.

[Ilovemathsmeth]

Well for something like a parabola, you could change to turning point form and THEN work out the coordinates of the turning point. You could also use the x = (-b)/2a rule too.

However for something like a cubic, you've got to use calculus. Or calculator