Questions on Addition of Ordinates

[darkphoenix]

1. State the conditions on k such that the graph of has

a) two vertical asymptotes
b) only one vertical asymptote
c) no vertical asymptotes

2. State the conditions on k such that the graph of has

a) two vertical asymptotes
b) only one vertical asymptote
c) no vertical asymptotes

3. If the curve has a vertical asymptoe at x=4 and has a horizontal asymptote at y=2 and a turning point at (-3,0) then find the values of A, B, b and c

[the.watchman]

1) Relating to ,

So for two asymptotes,

One asymptote,

No asymptotes,



2) Relating to ,

So for two asymptotes,

One asymptote,

No asymptotes,



3) Horizontal asymptote at y=2,

Vertical asymptote at x=4, ,

Other vertical asymptote at x=-10, because of symmetry about turning point at x=-3

[Mao]

3.
By inspection, the horizontal asymptote indicates B=2

By symmetry, the turning point is the mirror axis. Hence, the other vertical asymptote is at x=-10. The quadratic in the denominator is

Then, substituting the point (-3,0) gives

[the.watchman]

I got A=-98, how did I go wrong Mao?

[TrueTears]

from ur working u didnt

[the.watchman]

from ur working u didnt

Lol, yay! Let's hope this good luck streak carries over to spesh next year...

[TrueTears]

ok.

[Mao]

I got A=-98, how did I go wrong Mao?

By making an assumption:

Relating to ,

If the discriminant is 0, there will only be one v. asymptote, and there wouldn't be a turning point. The point (-3,0) in your graph would be an x-intercept. In this case, "turning point" indicates there are two vertical asymptotes. This relates to knowing the family of graphs and their shapes.

Your working out are correct, but it is important to not make unwarranted assumptions. I have recently made a very similar mistake in a project I'm working on, an incorrect assumption that cost'd me two days intensive of calculations.

[the.watchman]

I got A=-98, how did I go wrong Mao?

By making an assumption:

Relating to ,

If the discriminant is 0, there will only be one v. asymptote, and there wouldn't be a turning point. The point (-3,0) in your graph would be an x-intercept.

Ah I see, how should I change my working then?

[Mao]

I got A=-98, how did I go wrong Mao?

By making an assumption:

Relating to ,

If the discriminant is 0, there will only be one v. asymptote, and there wouldn't be a turning point. The point (-3,0) in your graph would be an x-intercept.

Ah I see, how should I change my working then?

Firstly, ignore that assumption.

And then, there are two ways:

1. use symmetry like I did to obtain the value of the second root, then expand.

2. use calculus:

Let the denominator be

We already know

We also know the turning point has an x coordinate of -3, that is, [this is inferred since the turning point of the quadratic and the turning point of the reciprocal graph have the same x value]



The rest is trivial.

[the.watchman]

Ok thanks, I'll take note of this for next year then!

[Mao]

Ok thanks, I'll take note of this for next year then!

Just realised you are not a specialist student [yet]. Kudos on the efforts!

[the.watchman]

Ok thanks, I'll take note of this for next year then!

Just realised you are not a specialist student [yet]. Kudos on the efforts!

Lol, thanks

[darkphoenix]

Thanks guys, But

The answers in the back of the book are different...

1. a) (-infinity,-8) U (8,infinity)
b)  k=+ and - 8
c) (-8,8)

2. Same as question 1, except instead of 8 its 20

[the.watchman]

My answers are exactly the same, except I just haven't worked it to the end (eg. k^2-64>0 becomes k<-8 OR k>8 anyway)