Author Topic: questions XD (Read 1809 times) Tweet Share

hyunah · « **on:** April 30, 2014, 01:09:06 pm »

if some one could help me... it is very much appreciated thank you

given that f(x) = 1/ax^2 +bx +c where a<0 and b^2 = 4ac how does that graph look like?

keltingmeith · « **Reply #1 on:** April 30, 2014, 01:33:16 pm »

Firstly, is f(x) = 1/(ax^2) + bx + c or 1/(ax^2 + bx + c)?

hyunah · « **Reply #2 on:** April 30, 2014, 01:58:54 pm »

1/(ax^2 + bx + c) sorry

kinslayer · « **Reply #3 on:** April 30, 2014, 02:09:59 pm »

If $b^2 = 4ac$ then the discriminant $\Delta = b^2 - 4ac = 0$ and therefore $g(x) = ax^2 + bx + c = 0$ will have exactly 1 solution, at $x = -\frac{b}{2a}$ .

If $a < 0$ then $\lim_{x\to \pm \infty} g(x) = -\infty$ .

With that information you should be able to draw a picture of the graph of $g(x)$ . It's basically an upside down parabola with a turning point on the x-axis.

Now if $f(x) = \frac{1}{g(x)} = \frac{1}{ax^2 + bx + c}$ , then $f(x)$ has a vertical asymptote at $x = -\frac{b}{2a}$ and horizontal asymptote $f(x) = 0$ . It looks pretty much like an upside down version of the graph of $\frac{1}{x^2}$ which isn't surprising since the behaviour of the function is dominated by the $ax^2$ term.

Clicky for graph:

https://www.desmos.com/calculator/m9vujyrmjs

Alwin · « **Reply #4 on:** April 30, 2014, 02:13:16 pm »

hi

so first up, lets consider ax² + bx + c.
1. what does the b²=4ac mean?

answer

The discriminant is zero

2. are there any x-ints? (they will become asymptotes

answer

Yes, but only one since discriminant is zero. Note that it will be x = -b/2a

3. what does a<0 mean?

answer

upside down parabola

now we can guess at what 1/(ax² + bx + c) will be like and we just use the general approach for reciprocal graphs

eg at x-ints there are asymptotes, when |f(x)|=1 then |1/f(x)|=1, when |f(x)|<1 then |1/f(x)|>1, when |f(x)|>1 then |1/f(x)|<1 etc etc

hope it helps

EDIT: beaten

hyunah · « **Reply #5 on:** April 30, 2014, 04:47:55 pm »

thank you!!! i get it now

hyunah · « **Reply #6 on:** April 30, 2014, 04:50:44 pm »

what about this one please?
let z1 = cis(pi/4)
Write down the complex equation of the straight line which passess through the points z1 and -z1 in terms of conjugate z1?

this question was on the VCAA 2006 exam 2 q5aii)

kinslayer · « **Reply #7 on:** April 30, 2014, 06:11:42 pm »

Quote from: hyunah on April 30, 2014, 04:50:44 pm

what about this one please?
let z1 = cis(pi/4)
Write down the complex equation of the straight line which passess through the points z1 and -z1 in terms of conjugate z1?

this question was on the VCAA 2006 exam 2 q5aii)

The set of points along the line from $z_1$ and $-z_1$ is the set of points equidistant from both $\overline{z_1}$ and $-\overline{z_1}$ . This is because we have a situation where the four points $z_1, -z_1, \overline{z_1}, -\overline{z_1}$ form a square (in general this is not true).

The equation is: $|z + \overline{z_1}| = |z-\overline{z_1}|$

I have attached a picture which may help.

hyunah · « **Reply #8 on:** April 30, 2014, 06:19:06 pm »

i'm sorry can you please explain further?
i'm sorry i really stupid ...

kinslayer · « **Reply #9 on:** April 30, 2014, 06:42:58 pm »

Quote from: hyunah on April 30, 2014, 06:19:06 pm

i'm sorry can you please explain further?
i'm sorry i really stupid ...

No problem, this is not easy and I probably haven't explained it very well.

In the diagram I drew, the line joining $\overline{z_1}$ to $-\overline{z_1}$ makes a right angle where it crosses the line from $-z_1$ to $z_1$ . That means that the distance from the line to $\overline{z_1}$ is the same as the distance from the line to $-\overline{z_1}$ .

The same thing is true for any other point along that line. So if $z$ is on the line, then $|z - (-\overline{z_1})| = |z - \overline{z_1}|$ .

hyunah · « **Reply #10 on:** April 30, 2014, 08:06:20 pm »

thank you thank you so much...
your explainations are great!!!

hyunah · « **Reply #11 on:** April 30, 2014, 08:06:41 pm »

i have another vcaa question.

vcaa 2010 spesh exam 2 question 5d)

sorry and thank you and please

Search the forums now!

We have moved!

Author Topic: questions XD (Read 1809 times) Tweet Share

hyunah

questions XD

keltingmeith

Re: questions XD

hyunah

Re: questions XD

kinslayer

Re: questions XD

Alwin

Re: questions XD

hyunah

Re: questions XD

hyunah

Re: questions XD

kinslayer

Re: questions XD

hyunah

Re: questions XD

kinslayer

Re: questions XD

hyunah

Re: questions XD

hyunah

Re: questions XD

Recent Posts

User:
Password: