Make a conjecture about y = ax+b

[5ymmetrica1]

Homework Statement

I have to make a conjecture about y = ax+b in terms of the ratio of the x coordinate in regards to the y-coordinate of the function.

Homework Equations

y = ax + b

The Attempt at a Solution

So I need to investigate the function like this:

y = ax + b

f(x)=x2+y2

f(x)=x2+(ax+b)(ax+b)

f(x)=x2+a2 x2+2axb+b2

But I'm fairly sure need to get a result that the ratio x/y is equal to a/1
But I usually end up at x/y = 1/a+b
when using the method I've written above.

 

[tiny-tim]

Hi 5ymmetrica1!

But I usually end up at x/y = 1/a+b

How do you get that?

(and what does r², = x² + y², have to do with it anyway??)

 

[mfb]

Try to look at (two or maybe more) y-values for different x-values, and see how their differences are.

 

[5ymmetrica1]

Hi 5ymmetrica1!


How do you get that?

(and what does r², = x² + y², have to do with it anyway??)

because this relates to an earlier problem I posted and is investigated in the same way which was answered with this

Using Calculus:

f(x) = x² + y², where y = -2x + 5

so, f(x) = 5x² - 20x + 25;
therefore, f'(x) = 10x - 20.

If f'(x) = 0 when f(x) is minimum, then

0 = 10x - 20, so x = 2.

Then, y = y(x) = y(2) = -2(2) + 5 = 1.

The coordinate is (2, 1).


NOTE: This agrees with the other "non-Calculus" method I used earlier.

so I have to use this
f(x) = x² + y² where y = ax+b

so by expanding and differentiating I have tried several ways and seem to always get to 1/ a+b

but I've noticed that the ratio of the x coordinate/ y coordinate is usually x:1
and this value of x is in most cases the same as the value of 'a'

For example in the function y = -2x+5
the value of x = 2 and y =1
the ratio is 2:1

and I checked with my teacher who said that by doing the same with y= ax+b
I should get to something similar such as a/1

I hope someone can help me here as I'm completely lost and have been stuck on this problem for nearly a week.

 

[tiny-tim]

because this relates to an earlier problem I posted …

But that problem asked you to find the position that is closest to the origin (ie to minimise x² + y²).

This question doesn't, does it?

What exactly is the question?​

 

[5ymmetrica1]

I have to observe what I noticed when investigating the original function and make a conjecture about it.

Then I'm required to prove this conjecture with algebra.

The only thing I noticed was that in the ratio of x:y, the number of 'x' was usually equal to 'a' in the equation ax+b

So for the function -2x+5 has a ratio of 2:1 since the 'a' value is 2
and for the function -4x+7 has a ratio of 4:1 since the 'a' value is 4
ect.

So by proving this with algebra I need to get a result of a:1

 

[haruspex]

The only thing I noticed was that in the ratio of x:y, the number of 'x' was usually equal to 'a' in the equation ax+b

You mean that it was approximately equal, I hope.
Under what circumstances (x values) does the accuracy improve? Does this suggest anything about a trend?

 

[tiny-tim]

But that problem asked you to find the position that is closest to the origin (ie to minimise x² + y²).

This question doesn't, does it?

What exactly is the question?​

ohh, i think i see what the question is now …

are you asked to find the slope of the shortest line from the origin to y = ax + b ?

 

[5ymmetrica1]

ohh, i think i see what the question is now …

are you asked to find the slope of the shortest line from the origin to y = ax + b ?

Yes exactly, but I also need to relate this back to the original function, this question is basically asking.

What did you notice, make a conjecture about it.
Then prove this conjecture with algebra.

To do this I need the slope of the shortest line of y = ax+b

 

[tiny-tim]

so I have to use this
f(x) = x² + y² where y = ax+b

so by expanding and differentiating I have tried several ways and seem to always get to 1/ a+b

i've done it (now i understand the question), and i don't get that

start by differentiating the distance-squared, and finding the value of x for which that is the minimum,

then put that value of x into ax + b to find y

then find y/x ​

show us what you get

 

[5ymmetrica1]

Do you mean like this?


f(x)=x²+a² x²+2axb+b²

f '(x) = 2x + 2a(2x) + 2

2x + 2a(2x) + 2 = 0

2x + 2a(2x) = -2

2x = -2 - 2a(2x)

x = -1 - ax

y = a(-1 -ax) + b

 

[tiny-tim]

Hi 5ymmetrica1!

Yes, exactly like that!

Except, your algebra is completely up the spout …

f(x)=x²+a² x²+2axb+b²

f '(x) = 2x + 2a(2x) + 2

no, that should be 2x + a²(2x) + 2ab

2x = -2 - 2a(2x)
x = -1 - ax

nooo !

try again ​

 

[5ymmetrica1]

Thanks Tim, I've been away from math for a while so my algebra is pretty rusty

so I'm now getting

2x + a²(2x) +2ab = 0

2x + a²(2x) = -2ab

2x = -2ab - a²(2x)

x = -ab - a²x

is this correct?

 

[tiny-tim]

(just got up :zzz:)

better!

but don't you need x on the left and a constant on the right?

 

[5ymmetrica1]

So you mean I need all the x's on the left and the constant a^2 on the right?

 

[tiny-tim]

a² ?

no ​

 

[5ymmetrica1]

well I know 'b' was a constant

or do you mean I need to add a constant 'c' like in anti-differentiation?

I'm confused as well :P

 

[5ymmetrica1]

so I now know I need to get a final result of -a, which proves my original hypothesis that the value of the ratio is equal to the value of a in the equation y = ax+b

and it involves using the equation √x² + a²x²+2axb+b²

but I'm stuck from here. Can anyone offer me any help here. This assignment is due tomorrow and I'm still stuck on this problem!

 

[CompuChip]

x = -ab - a²x

is this correct?

If the letters confuse you, maybe it's useful setting the constants to a numeric value (e.g. a = 2, b = 3) and first working out the answer with those example values. I.e. maybe try solving
$$x = -2 \cdot 3 - 2^2 x$$
first.

 

[vela]

Thanks Tim, I've been away from math for a while so my algebra is pretty rusty

so I'm now getting

2x + a²(2x) +2ab = 0

2x + a²(2x) = -2ab[/color]

2x = -2ab - a²(2x)

x = -ab[/color] - a²x

is this correct?

You were close here. As Tim said, you need to get the x's all on one side and the constants on the other. Only the red term doesn't depend on x. It's the constant, not a², you want on the righthand side.

Go back to your second equation (the one in blue) because it has all the terms with x on the lefthand side and the constant on the righthand side. Divide out the common factor of 2, and then try solve for x again.

 

[5ymmetrica1]

You mean like this?

x=−2⋅3−2 2 x

-2(3) -2²x = x

-2(3) = 2²(2x)

-6 = 4(2x)

8x = -6

x = -6/8 = -3/4

 

[5ymmetrica1]

You were close here. As Tim said, you need to get the x's all on one side and the constants on the other. Only the red term doesn't depend on x. It's the constant, not a², you want on the righthand side.

Go back to your second equation (the one in blue) because it has all the terms with x on the lefthand side and the constant on the righthand side. Divide out the common factor of 2, and then try solve for x again.

so I get

x + a²x = -ab

a²(2x) = -ab

2x = -ab / a² = -b/a

x = (-b/a)/2

 

[vela]

No, that's not right. $x+a^2x$ isn't equal to $a^2(2x)$. For instance, if a=2, you'd have $x+a^2x = x+4x = 5x$ whereas $a^2(2x) = 8x$.

When you add, say, x and 3x, you get 4x. How do you come up with the 4? You add the coefficients. In your problem, a² is part of the coefficient.

 

[5ymmetrica1]

so x + a^2(x) would be 2a^2(x)?

 

[vela]

No. That's no different than what you wrote before. What's the coefficient of the first term? Of the second term? What do you get when you add them together?

 

[5ymmetrica1]

I know how to do it with numbers but the a^2 is confusing me, if I have a^2 amounts of (x) and I add (x)
x is the coefficient in the first and second term isn't it? if I add x and x together I have 2x

 

[vela]

No, the coefficient is everything but the x.

 

[5ymmetrica1]

so x + a^2(x) = a^2(x)
I hope so anyway or I'm going to bury my head in the sand

 

[tiny-tim]

Hi 5ymmetrica1!

so x + a^2(x) would be 2a^2(x)?

I know how to do it with numbers but the a^2 is confusing me, if I have a^2 amounts of (x) and I add (x)
x is the coefficient in the first and second term isn't it? if I add x and x together I have 2x

hmm … you're really bad at this algebra thing, aren't you?

x + a²x is the same as 1(x) + a²(x)

forget the word "coefficient"

look up the "distributive law"

 

[5ymmetrica1]

Hi 5ymmetrica1!



hmm … you're really bad at this algebra thing, aren't you?

x + a²x is the same as 1(x) + a²(x)

forget the word "coefficient"

look up the "distributive law"

Actually I've been away from my studies for a long time due to some health issues and have been struggling to get myself back to where I was 6 months ago, I have finals coming up too which is worrying!

Is my last post correct?