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Quadratics from turning point - any maths teachers about?

(6 Posts)
bandito Mon 30-Apr-18 20:01:29

I'm not bad at maths but haven't touched quadratics since A Level a VERY long time ago. Poor DD is doing them for GCSE (so much harder than it used to be but that's another thread).

She has a quadratic question which asks for the values of a and b where the curve has a turning point of (-3,-4) where a and b are integers and y=x(squared - can't find symbol,sorry) + ax +b.

I am a bit stumped how to do this - any ideas?

OP’s posts: |
noblegiraffe Mon 30-Apr-18 21:42:34

A quadratic in completed square form, (x+a)^2+b has the turning point (-a,b).
So your quadratic is (x+3)^2-4

Then just expand the brackets to get it into the required form.

tweetypi Mon 30-Apr-18 21:45:00

This is a transformations of functions question. If the turning point has moved from (0,0) to (-3,-4), the original graph of y=x2 has moved three left and 4 down, so has become y=(x+3)^2 - 4

Then you'd need to expand and get it into the form the q asked for.

xsquared Mon 30-Apr-18 21:49:22

Completing the square of x^2 +ax+b, and express as (x+p)^2 +q will give you the co-ordinates of the midpoint of the curve, (-p,q). The co-ordinates represent the translation of the point (0,0) of the curve f(x) = x^2 under the function f(x+2)+q

So 2p is the coefficient of x, i.e 2p=a and
p^2 + q = b, the constant

In this case,
p= 3 and q = -4, so
6=a and
3x3 -4 = 5 = b

xsquared Mon 30-Apr-18 21:50:23

sorry, that should be f(x+p)+q

bandito Tue 01-May-18 20:31:51

Thank you!
A quadratic in completed square form, (x+a)^2+b has the turning point (-a,b).
This I did not know but DD needs to learn and then they are quite straight forward I think. It was only a 3 mark question so I knew I was missing something!

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