Maths Help Please

Trigger · 20 Feb 2006 at 15:48

Hi,

Would someone mind helping me with a maths question I have? It's not a test or anything, it's just something our teacher has given us over the holidays to do. I've looked in my textbook and found nothing on the subject and have searched google only to find someone trying to flog me a programme

Basically, the question is:

Factorise y=2x² - 7x -15 which I think I have done fine and got:

(2x + 3)(x - 5)

Then...

The graph of y=2x² - 7x -15 is sketched below:

(Excuse dodgey drawing

)

Find the equation of the line of symmetry of this graph...

Anyone have any ideas? It's probably really simple but we haven't covered it yet so I would appreciate any help

Thanks

Ben

anksta · 20 Feb 2006 at 15:50

its half way between P and Q, like if P was -4 and Q was 6 then the equation of that line would be x=1

Trigger · 20 Feb 2006 at 15:52

anksta said:
its half way between P and Q, like if P was -4 and Q was 6 then the equation of that line would be x=1

Yeah, but there are no values on the axis i've been given- they're just blank and I can't just write 'halfway between P and Q' as the answer lol

Thanks for the reply BTW

Ben

KaHn · 20 Feb 2006 at 15:52

Yeah as anksta has said,

You need to find the points where y=0

so since youve factorised it you should be almost there.

KaHn

Trigger · 20 Feb 2006 at 15:54

KaHn said:
Yeah as anksta has said,

You need to find the points where y=0

so since youve factorised it you should be almost there.

KaHn

Ahhh, I think I understand now. There should be two posible values for y=0 and the line of symmetry will be halfway in between them?

Thanks again

Ben

KaHn · 20 Feb 2006 at 15:56

Trigger said:
Ahhh, I think I understand now. There should be two posible values for y=0 and the line of symmetry will be halfway in between them?

Thanks again

Ben

Yes basically equate both brackets to zero and work out them both seperatally for the two values of x.

KaHn

robmiller · 20 Feb 2006 at 15:56

That factorisation isn't right, is it?

KaHn · 20 Feb 2006 at 15:57

robmiller said:
That factorisation isn't right, is it?

I didnt think so until i checked it with paper :rolleyes:

been a while since ive done GCSE

Trigger · 20 Feb 2006 at 15:58

KaHn said:
/edit:- also your factorisation is wrong

How exactly?

y=2x² - 7x -15 --> (2x + 3)(x - 5)

If I multiply the bit in the brackets out I get what I started with :confused:

2x multiplied by x = 2x²

3 multiplied by x = 3x

-5 multiplied by 2x = -10x

and +3 multiplied by -5 = -15, so...

2x² + 3x -10x -15 = 2x² -7x -15 doesn't it? :confused:

KaHn · 20 Feb 2006 at 15:58

Trigger said:
How exactly?

y=2x² - 7x -15 --> (2x + 3)(x - 5)

If I multiply the bit in the brackets out I get what I started with

2x multiplied by x = 2x²

3 multiplied by x = 3x

-5 multiplied by 2x = -10x

and +3 multiplied by -5 = -15, so...

2x² + 3x -10x -15 = 2x² -7x -15 doesn't it?

see my ninja edit

I looked at it quickly and weny b^2-4ac<0 hence imaginary but then i remembered the minus sign

KaHn

gord · 20 Feb 2006 at 16:03

Thats the problem with asking any question on OcUK.. even if youve got one bit right already it can come into question..

Trigger · 20 Feb 2006 at 16:06

KaHn said:
see my ninja edit

I looked at it quickly and weny b^2-4ac<0 hence imaginary but then i remembered the minus sign

KaHn

hmmm, damn my slow keyboard skillz

Sorry

So would I be right in saying that the line of symmetry is x = -1?

Thanks

Ben

spirit · 20 Feb 2006 at 16:08

y=2x² - 7x -15
y=2(x² - 3.5x) - 15
y=2(x-1.75)² - 18 1/16

thus x=1.75 ?

KaHn · 20 Feb 2006 at 16:08

Trigger said:
hmmm, damn my slow keyboard skillz Sorry So would I be right in saying that the line of symmetry is x = -1?

Thanks

Ben

No that is wrong.

You need to look a (2x+3)=0 and (x-5)=0

Then basically from those two equations what does x=?

KaHn

KaHn · 20 Feb 2006 at 16:09

spirit said:
y=2x² - 7x -15

(2x - 3.5)^2 - 27.25

hence the line of symetry it gonna be 3.5/2 = 1.75

much easier to complete the square to find symetrys for quadratics.

Look at how the question is worded, factorise this then get the line of symmetry.

KaHn

spirit · 20 Feb 2006 at 16:09

..edited out something some already said, but i think my way is faster, eventho i made a stupid mistake first time round

spirit · 20 Feb 2006 at 16:11

KaHn said:
Thats wrong but nevermind.

Look at how the question is worded, factorise this then get the line of symmetry.

KaHn

i see where i made my mistake lol, i did this long time ago

Trigger · 20 Feb 2006 at 16:11

KaHn said:
No that is wrong.

You need to look a (2x+3)=0 and (x-5)=0

Then basically from those two equations what does x=?

KaHn

Well that's what I did but wasn't sure whether it was x = 3 and x = -5 or if the 2x in (2x+3) had an effect on the 3 or not? But wouldn't that make it x = 1.5 and x = -5 ?

Thanks

Ben

gord · 20 Feb 2006 at 16:12

KaHn said:
No that is wrong.

You need to look a (2x+3)=0 and (x-5)=0

Then basically from those two equations what does x=?

KaHn

meh

KaHn · 20 Feb 2006 at 16:12

spirit said:
i see where i made my mistake lol, i did this long time ago

Sorry I thought you were doing the locations of x and not the line

KaHn