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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: cheunga on October 14, 2017, 05:25:39 pm

Title: Yr.11 Stationary Points
Post by: cheunga on October 14, 2017, 05:25:39 pm
Hi, can someone please help me with this question, bearing in mind that this is at Yr.11 level so as dumbed down as possible a solution would be appreciated!  Thanks in advance.
The curve with equation y=ax2+bx+c has a stationary point at (1,2).  When x=0, the slope of the curve is 45o.  Find a,b and c.
Title: Re: Yr.11 Stationary Points
Post by: Eric11267 on October 14, 2017, 05:35:04 pm
Quote from: cheunga on October 14, 2017, 05:25:39 pm
Hi, can someone please help me with this question, bearing in mind that this is at Yr.11 level so as dumbed down as possible a solution would be appreciated!  Thanks in advance.
The curve with equation y=ax2+bx+c has a stationary point at (1,2).  When x=0, the slope of the curve is 45o.  Find a,b and c.
Okay so differentiating it gives you 2ax+b
At x=0 the slope is 45 degrees, which is a gradient of 1
So we have 2*a*(0)+b=1
b=1
Also at x=1 there is a turning point so the derivative equals 0
So we have 2a+1=0
a=-1/2
Finally putting the values of a and b back into the original equation and subbing in (1,2) gives c=3/2
Title: Re: Yr.11 Stationary Points
Post by: cheunga on October 14, 2017, 05:49:25 pm
Hey Eric11267, thanks so much for your help! Much appreciated.
Title: Re: Yr.11 Stationary Points
Post by: Sigma on October 19, 2017, 05:39:19 pm
Hey Eric 11267,

Thank you so much for your help. I was stuck on this question too. I also find these questions the most difficult. Dy/x a function and then subsiding in x to find y and thus giving you the coordinates for the local max and local min is easy - it is just these questions are always hard to answer. Finding 3 variables is hard considering it is only worth 3 marks. The first variable easy, it is the last two that are hard.
Title: Re: Yr.11 Stationary Points
Post by: Yueni on October 23, 2017, 07:04:50 pm
I'd like to chime in here and say that the functional form you are given isn't necessarily the form you must do your working out in.

You are given a turning point, why not use turning point form for your quadratic? Then derive that and substitute x=0, dy/dx=1.
You can then expand your final quadratic to get a, b and c.

You save a little bit of time! And time is all so important when you never have enough of it.
Title: Re: Yr.11 Stationary Points
Post by: Sigma on October 24, 2017, 04:32:52 pm
thanks for your help.

Eric11267's way is how the worked solutions do it and my teacher does it so I'll stick with that method. Thanks tho.