Maths Help Please

KaHn · 20 Feb 2006 at 16:14

gord said:
so your saying 3.25 is the line of symmetry?

No im saying that 3.25 is the middle of that parabolic curve, so that is 3.25 further along from x=-1.5 would be the middle so the line of symmetry is x=1.75

KaHn

Ricochet J · 20 Feb 2006 at 16:15

Factorisation and drawing graphics for quadtratic equations is dead easy if you know how to complete the square.

Assuming you do, if you complete the square of ax^2+bx+c then you'll get all the values you need to plot a graph and you'll get the solutions

gord · 20 Feb 2006 at 16:16

KaHn said:
No im saying that 3.25 is the middle of that parabolic curve, so that is 3.25 further along from x=-1.5 would be the middle so the line of symmetry is x=1.75

KaHn

Yeh sorry, i didnt explain mine clearly.. i make it 1.75 too

Trigger · 20 Feb 2006 at 16:17

KaHn said:
No im saying that 3.25 is the middle of that parabolic curve, so that is 3.25 further along from x=-1.5 would be the middle so the line of symmetry is x=1.75

KaHn

Ahh, I see now. It was because I was not dividing by 2 before in (2x+3) when I should have done so

Anyways, thank you to everyone for their help, you've all been great

Thanks again :cool:

Ben

spirit · 20 Feb 2006 at 16:22

a rediculously fast way to do ** maths homework, post it on ocuk forums :}

Psiko · 20 Feb 2006 at 16:25

Notice also that if we complete the square (or differentiate) we find that x = 7/4 corresponds to a minimum. Since there is only one minimum it must be ON the axis of symmetry. We also know that to go from the graph of y = x^2 (which is symmetric in x=0) to your parabola we only need to do translations and stretches so the line of symmetry must stay vertical, hence it must be x = 7/4. This is probably more information than you need to know. For this question you can just use intuition really.

daz · 20 Feb 2006 at 16:27

Edit: doesn't look like you need to bother with finding the turning point...

You could have just differentiated twice.

Ricochet J · 20 Feb 2006 at 16:33

daz said:
Edit: doesn't look like you need to bother with finding the turning point...

You could have just differentiated twice.

I'm curious to how that may be.

dy/dx would give you gradient
d^2y/dx^2 would give you the minimum and turning point and weather its an increasing or decreasing funtion.

Do you say this because d^2y/dx^2 will provide a minimum and turning point which will relate to x= the value of what you just differentiated twice?

spirit · 20 Feb 2006 at 16:35

daz said:
Edit: doesn't look like you need to bother with finding the turning point...

You could have just differentiated twice.

You only need to differentiate once, it is completely irrelevant whether it is a maximum or minimum (or inflection) in this case as it it an X2 graph.

Ricochet J · 20 Feb 2006 at 16:37

spirit said:
You only need to differentiate once, it is completely irrelevant whether it is a maximum or minimum (or inflection) in this case as it it an X2 graph.

Aigh, you speak the truth. I reckon if he differentiatied once and made the gradient function equal zero then it would've worked as well.

spirit · 20 Feb 2006 at 16:37

eXSBass said:
I'm curious to how that may be.

dy/dx would give you gradient
d^2y/dx^2 would give you the minimum and turning point and weather its an increasing or decreasing funtion.

Do you say this because d^2y/dx^2 will provide a minimum and turning point which will relate to x= the value of what you just differentiated twice?

It will d2y/dx2 will tell you what type of turning point it is, or help you figure it out. if its positive its a minimum, if its negative its a maximum, if its zero u need to check it

daz · 20 Feb 2006 at 16:41

You only need to differentiate once, it is completely irrelevant whether it is a maximum or minimum (or inflection) in this case as it it an X2 graph.

True, I had a brain fart: there's no real need to differentiate twice.

Been doing bloody fourier transforms all morning.

callmeBadger · 20 Feb 2006 at 16:44

what was the formula that would find y=0, its something like -b +-((b^2-4ac)^-2)/2a or something. can't remeber what the equation was called.

Ohh how the greyt matter hurts for a slow monday at work.

daz · 20 Feb 2006 at 16:50

I prefer to "complete the square" rather than use the quadratic formula.

Treefrog · 20 Feb 2006 at 16:56

eXSBass said:
Aigh, you speak the truth. I reckon if he differentiatied once and made the gradient function equal zero then it would've worked as well.

Yup. x=7/4=1.75, same as averaging -1.5 and +5. Always best to double check.

Ricochet J · 20 Feb 2006 at 16:58

daz said:
I prefer to "complete the square" rather than use the quadratic formula.

Completing is using the quadtratic formula but very subtly. When you do complete the square of a quadtratic you are in actual fact completing the square of ax^2+bx+c=0. The difference is a, b and c are numbers

If you dont believe me complete the square of ax^2+bx+c=0

Badger: Its x=(-b±"square root"b²-4ac)/2a

Arcade Fire · 20 Feb 2006 at 17:00

I love how these threads always get more complicated than they need to - how did we get onto differentiation?!

callmeBadger · 20 Feb 2006 at 17:04

knew it was something like that, after 5 mins on excel remembered formula, was not sure where to put brackets, ahh well, just got to find the ASCII codes for the square root sign and +- sign then my afternoon of work has actually been useful for once.

Ricochet J · 20 Feb 2006 at 17:04

Ahahaha, so true

daz · 20 Feb 2006 at 17:10

eXSBass said:
Completing is using the quadtratic formula but very subtly. When you do complete the square of a quadtratic you are in actual fact completing the square of ax^2+bx+c=0. The difference is a, b and c are numbers

If you dont believe me complete the square of ax^2+bx+c=0

Badger: Its x=(-b±"square root"b²-4ac)/2a

It's easier to remember the method of completing the square rather than the quadratic formula, is what I was alluding to.

Arcade Fire said:
I love how these threads always get more complicated than they need to - how did we get onto differentiation?!

We're finding turning points!