Maths Help

[monaro_man_91]

Ok I've got the zeros, y-intercept and vertex of a quadratic equation, now I just need to form the quadratic equation.

Zeros - (96,0) and (-96,0)
Y-Intercept - (0,192)
Vertex - (0,192)

How do I get the quadratic equation from this? The best I can get is y = ax^2 + by + 192. Is it even possible to get the quadratic equation for it?

Here is the question if you want it:

Many architectural shapes are parabolic or approximately parabolic. For example, the stainless steel Gateway Arch at St Louis, Missouri, USA, has a span (width of base) of 192m and a height of 192m. Find an algebraic equation to describe the curve.

Thanks.

[Maroon]

1 + 1 = Window

[monaro_man_91]

1 + 1 = Window

Thanks mate! I knew it, I just had a mental blank! :p

But seriously, thanks but no thanks.

[n00b]

My head doesn't hurt.

[Broncos v cowboys]

How do I get the quadratic equation from this? The best I can get is y = ax^2 + by + 192. Is it even possible to get the quadratic equation for it?





Thanks.

And a-hoobie-joo to you too.

[spano]

Just Sub the information into the equation y = ax2 + bx + c

With the y-intercept (0,192) 192 = a02 + b0 + c
therefore 192 = 0 + 0 + c
therefore 192 = c
Now you got y = ax2 + bx + 192

Next you use the zeros (96,0) and (-96,0)
Therefore 0 = a(96)(96) + b(96) + 192
and 0 = a(-96)(-96) + b(-96) +192
therefore 0 = 9216a + 96b + 192
and 0 = 9216a - 96b + 192
and now you sub the equations together
and you get
0 = 18432a + 384
this done by adding up the like terms in each equation together
this causes +96b +(-96b) to cancel out
Just rearrange the equation and you get
-18432a = 384
to get what a equals divide each side by -18432
and you get
a = 384/-18432
therefore a = -1/48
since the b cancels out it eqauls zero you can test this out by sub back into the equation
0 = 9216(-1/48) + 96b + 192
therefore 0 = -192 + 96b + 192
therefore 0 = 96b
therefore b = 0/96
therefore b = 0
Now you get your equation: y = -1/48x2 + 192

[raiders#1 2006]

Just Sub the information into the equation y = ax2 + bx + c

With the y-intercept (0,192) 192 = a02 + b0 + c
therefore 192 = 0 + 0 + c
therefore 192 = c
Now you got y = ax2 + bx + 192

Next you use the zeros (96,0) and (-96,0)
Therefore 0 = a(96)(96) + b(96) + 192
and 0 = a(-96)(-96) + b(-96) +192
therefore 0 = 9216a + 96b + 192
and 0 = 9216a - 96b + 192
and now you sub the equations together
and you get
0 = 18432a + 384
this done by adding up the like terms in each equation together
this causes +96b +(-96b) to cancel out
Just rearrange the equation and you get
-18432a = 384
to get what a equals divide each side by -18432
and you get
a = 384/-18432
therefore a = -1/48
since the b cancels out it eqauls zero you can test this out by sub back into the equation
0 = 9216(-1/48) + 96b + 192
therefore 0 = -192 + 96b + 192
therefore 0 = 96b
therefore b = 0/96
therefore b = 0
Now you get your equation: y = -1/48x2 + 192
W.T.F.

*Leaves*

[League Lover]

w0w spano.. just w0w!

[BrisbaneRabbitohs]

im doing maths at uni and ive got no idea wat that was about, are zeros where it cuts the x axis? i think ill give it a try if thats so.

[spano]

Yes zeros are where it cuts the x-axis BrisbaneRabbitohs

[League Lover]

are you a genuis by any chance spano?

[spano]

are you a genuis by any chance spano?
No i'm not i'm just alright maths i suppose nothing else. except swimming

[League Lover]

:) im in the top math class at school but my teacher gives us lyk grade 10 work when were only in grade 8 :eek:

[freddie fittler]

:) im in the top math class at school but my teacher gives us lyk grade 10 work when were only in grade 8 :eek:

Serious!:eek:

That explains a lot.

[League Lover]

yea lol :(

[dondogz]

1+1=6 i think.......

[League Lover]

bout right tbh...

[GCPanther]

im doing maths at uni and ive got no idea wat that was about, are zeros where it cuts the x axis? i think ill give it a try if thats so.

yeah same with me, I do maths at uni and I'm lost with this stuff lol

[SIDHE-ROX!!]

Many architectural shapes are parabolic or approximately parabolic. For example, the stainless steel Gateway Arch at St Louis, Missouri, USA, has a span (width of base) of 192m and a height of 192m. Find an algebraic equation to describe the curve.

If you didn't understand Spano's explanation (I didnt), this might be a little easier to follow, but possibly wrong. lol

X-Intercepts = (0, 0) and (192, 0)
Y-Intercepts = (0, 0)
Turning Point = (96, 96)

y = Ax^2 + Bx + C

We know that when X = 0, Y = 0, so we substitute into the equation.

0 = 0A + 0B + C
therefore C = 0

We know that when X = 192, Y = 0

0 = 36864A + 192B

When X = 96, Y = 96.

therefore: 96 = 9216A + 96B

Now we have a simultaneous equation.

We make the variable of A equal to eachother so we can cancel the two out.
First, we make A the subject of the two equations.

1) 0 = 36864A + 192B
36864A = -192B

2) 96 = 9216A + 96B
9216A = 96-96B

Now we want the co-efficients of A to be the same.

36864/9216 = 4

therefore 36864A = 4(9216A)

Now we multiply the second equation by 4 to achieve identical equations.

4(9216A) = 4(96-96B)
36864A = 384 - 384B

AND 36864A = -192B

therefore:

-192B = 384 - 384B
192B = 384
B = 2

Now we substitute this into any equation with A in it.

96 = 9216A + 96B

But B = 2

96 = 9216A + 192
9216A = -96
A = -1/96

Now, substitute these values into the original equation.

y = Ax^2 + Bx + C
y = (-x^2)/96 + 2x

And thats your answer.

That was more difficult than I originally thought, hope its not too confusing. Just ask me if you dont understand anything.

[monaro_man_91]

Lol thanks for the help guys, I was studying for a maths exam when I was trying to figure this out. I did that exam a few weeks ago and totally pwned it, A's in each section, woot :D

[SIDHE-ROX!!]

Lol thanks for the help guys, I was studying for a maths exam when I was trying to figure this out. I did that exam a few weeks ago and totally pwned it, A's in each section, woot :D

ARGH!!! I should really look at dates, what a waste of time and thought! lol

btw, was I right? lol

[Suncorp Steve]

Maths A is where its at :D

[spano]

[QUOTE=SIDHE-ROX!!;712543]If you didn't understand Spano's explanation (I didnt), this might be a little easier to follow, but possibly wrong. lol

X-Intercepts = (0, 0) and (192, 0)
Y-Intercepts = (0, 0)
Turning Point = (96, 96)]

You did it right with the wrong values it should be

x-intercepts = (96,0) and (-96,0)
y-intercepts = (0,192)
Turning Point = (0,192)

and that's why we got a different answer

[SIDHE-ROX!!]

If you didn't understand Spano's explanation (I didnt), this might be a little easier to follow, but possibly wrong. lol

X-Intercepts = (0, 0) and (192, 0)
Y-Intercepts = (0, 0)
Turning Point = (96, 96)

You did it right with the wrong values it should be

x-intercepts = (96,0) and (-96,0)
y-intercepts = (0,192)
Turning Point = (0,192)

and that's why we got a different answer

AHH I see. The main reason we got different answers was because I got the turning point wrong, using my points, the turning point should have been (96, 192)

If I had have used that value, we would have gotten the same answer, just a different constant at the end.

[spano]

AHH I see. The main reason we got different answers was because I got the turning point wrong, using my points, the turning point should have been (96, 192)

If I had have used that value, we would have gotten the same answer, just a different constant at the end.
No the turning has to be (0,192). If you draw it on a graph it has a local maximum at (0,192). Your turning point is right above the x-intercept and therefore is not a parabola. Also your x-intercepts and y-intercept are wrong too. Look at the first post in the thread.

[League Lover]

spano and sidhe rox!!! i just made you a math thread.... so you can go discuss stuff in that :)

[SIDHE-ROX!!]

No the turning has to be (0,192). If you draw it on a graph it has a local maximum at (0,192). Your turning point is right above the x-intercept and therefore is not a parabola. Also your x-intercepts and y-intercept are wrong too. Look at the first post in the thread.

The way I did it was assuming that the bridge started at (0, 0), therefore, it ends at (192, 0) because its 192m in length.

If its a perfect parabola, then the turning point will be right in the centre of the two X co-ordinates, which is 0 and 192. The middle of that is 96, therefore, the X co-ordinate of the turning point is 96. The Y Co-ordinate was 192 as said in the question. Therefore, the turning point was (96, 192)

Correct?

If so, our parabola's should be exactly the same, just with a different C value. In fact, Ill work it out now.

[SIDHE-ROX!!]

X-Intercepts = (0, 0) and (192, 0)
Y-Intercepts = (0, 0)
Turning Point = (96, 192)

y = Ax^2 + Bx + C

We know that when X = 0, Y = 0, so we substitute into the equation.

0 = 0A + 0B + C
therefore C = 0

We know that when X = 192, Y = 0

0 = 36864A + 192B

When X = 96, Y = 192.

therefore: 192 = 9216A + 96B

Now we have a simultaneous equation.

We make the variable of A equal to eachother so we can cancel the two out.
First, we make A the subject of the two equations.

1) 0 = 36864A + 192B
36864A = -192B

2) 96 = 9216A + 96B
9216A = 192-96B

Now we want the co-efficients of A to be the same.

36864/9216 = 4

therefore 36864A = 4(9216A)

Now we multiply the second equation by 4 to achieve identical equations.

4(9216A) = 4(192-96B)
36864A = 768 - 384B

AND 36864A = -192B

therefore:

-192B = 768 - 384B
192B = 768
B = 4

Now we substitute this into any equation with A in it.

96 = 9216A + 96B

But B = 4

96 = 9216A + 384
9216A = -288
A = -1/32

Now, substitute these values into the original equation.

y = Ax^2 + Bx + C
y = (-x^2)/32 + 4x

Guess it is kinda different. Ill compare both graphs in graphmatica.

ok, my graph is weird, where have i gone wrong? Assuming it starts at (0, 0)

EDIT: The real answer should be: y = (-x^2)/48 + 4x

Ive tried back tracking, but it all doesn't make sense :confused:

[SIDHE-ROX!!]

GAH I cant edit.

Anyway I found the problem. There was one value I forgot to change from 96 to 192 when I did it before.

Now we substitute this into any equation with A in it.

192 = 9216A + 96B

But B = 4

192 = 9216A + 384
9216A = -192
A = -1/48

Now, substitute these values into the original equation.

y = Ax^2 + Bx + C
y = (-x^2)/48 + 4x

There we go, proven, apart from that error i did at the start, I was correct :D

[spano]

Here I'll have a go at your figures:
x-intercept (0,0) and (192,0)
y-intercept (0,0)
turning point (96,192)

y = ax^2 + bx + c
now sub (0,0)
therefore 0 = a0^2 + b0 + c
therefore 0 = 0 + 0 + c
therefore c = 0
now you get y = ax^2 + bx
sub (192,0) and (96,192) into the equation and sub equations

0 = a(192)^2 + 192b
and 192 = a(96)^2 + 96b

0 = 36864a + 192b (1)
and 192 = 9216a + 96b (2)
multiply eqaution 2 by -2
and you get

-384 = -18432 - 192b
add equations together

-384 = 18432a

therefore a = -384/18432
therefore a = -1/48

Sub a = -1/48 into one of the equations (1), (2) or (3) they all work equally and will give the same b value. I will use (1)

therefore 0 = -768 + 192b
therefore 768 = 192b
therefore b = 768/192
therefore b = 4

and you get y = -(x^2)48 + 4x

yes your right it is shifted to the right by +96 that is on the x-axis

[n00b]

OM*G!


When I look at these.. some sort of equations my mind goes nuts and my eyes go blurry. Deadset.

English > Math

[vincehellier]

My head hurts :confused: :confused: :confused: :confused: :confused: :confused: :confused: :confused: :confused: :confused: