# Curve Sketching

• December 19th 2008, 03:19 PM
Lonehwolf
Curve Sketching
These are A level past papers, and unluckily I can't grab anything out of these. Any and all help is appreciated. I need a complete rundown on each question, this is for tomorrow. The more I can understand (mind you I'd rather understand then have you solve them for me without getting anything out of it), the better.

1: A curve has the equation

$y=\frac{5}{x-5} - \frac {2}{x-2}$

(i) Find the equations as the three asymptotes.
(ii) Find the coordinates of the turning points of the curve.
(iii) Draw a sketch of the curve and hence or otherwise, determine the set of values of y for which no point of the curve exists.

2: As described in this picture below...
http://img67.imageshack.us/img67/466...age3av6.th.jpg

3: The graph of

$y=\frac{ax^2+bx+c}{x^2+qx+r}$ has the lines $x=1$, $x=3$ and $y=2$ as asymptotes and a turning point at $(0,1)$. Find the constants $a, b, c, q$ and $r$ and show that the graph has a second turning point.

Sketch the graph showing clearly its turning points and its behaviour as it approaches the asymptotes.

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I'm at a loss in here, and I simply can't move deep enough to go to any clear solution. I get stuck and rather than showing my half scrubbed working I'd like to see how you'd do it from start to end with some thorough explanations. All 3 seem pretty similar, so working any one of them, whichever you find best, would be a big help.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thanks,

Wolf.
• December 19th 2008, 05:32 PM
Mush
Quote:

Originally Posted by Lonehwolf
These are A level past papers, and unluckily I can't grab anything out of these. Any and all help is appreciated. I need a complete rundown on each question, this is for tomorrow. The more I can understand (mind you I'd rather understand then have you solve them for me without getting anything out of it), the better.

1: A curve has the equation

$y=\frac{5}{x-5} - \frac {2}{x-2}$

(i) Find the equations as the three asymptotes.

The vertical asymptotes of a curve are the points where the curve is undefined. A divide by zero gives an undefined point. Hence the undefined points here are when the denominator of either fraction is zero. What values of x does this occur for? The Horizontal asymptotes of a curve are given by the values that the curve converges towards as x approaches positive or negative infinity. A good way to do this is to put your equation of y into a single fraction ( $\frac{3x}{x^2-7x+10}$) and use long division to get an equation. Incidentally, then answer is $\frac{3}{x}...+ (a bunch of other terms)$. In general, the answer to your long division will be of the form $ax^2+bx+c+\frac{d}{x}$. A general rule of thumb is to let the curve of the asymptote be defined as $y_asymptote =$(all the terms up until the $\frac{d}{x}$ term. In our case, the term before the $\frac{d}{x}$ term is simply 0! Then you let the curve you're interested in by defined as the sum as all the terms INCLUDING the $\frac{d}{x}$. In this case $y_{curve} = \frac{3}{x}$. Then what you do is compare these two curves. How does ycurve approach yasymptote for positive and negative infinity?. Well for positive infinity, then ycurve is going to be very small and very positive. So the curve approaches 0 from ABOVE for positive infinity. For negative infinity, the ycurve will be very small and negative, meaning it approaches y = 0 from below! These are your asymptotes!

(ii) Find the coordinates of the turning points of the curve.

Differentiate the function, set the result = 0, solve for x, then plug these x values back into the original equation to get corresponding y values.

(iii) Draw a sketch of the curve and hence or otherwise, determine the set of values of y for which no point of the curve exists.

Use a table of sign to evaluate the values of the curve at important points in the curve (important points being turning points and asymptotes!)
2: As described in this picture below...
http://img67.imageshack.us/img67/466...age3av6.th.jpg

3: The graph of

$y=\frac{ax^2+bx+c}{x^2+qx+r}$ has the lines $x=1$, $x=3$ and $y=2$ as asymptotes and a turning point at $(0,1)$. Find the constants $a, b, c, q$ and $r$ and show that the graph has a second turning point.

Sketch the graph showing clearly its turning points and its behaviour as it approaches the asymptotes.

Well. The vertical asymptotes are given by x = 1 and x=3. This means, as I explained earlier, that the denominator is equal to zero when x =3 and x = 1. So sub these values in to the denominator, and set them equal to zero. You will then have two equations in q and r that can be solved simultaneously. Also, use long division and see if you can find any juicy information from the curve knowing that the horizontal asymptote is y = 2. Also, if you differentiate the function using the quotient rule, set it equal to zero (remember that only the denominator can be zero!) and then sub in the values for your turning point (0,1), then you should be able to solve for a and b.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I'm at a loss in here, and I simply can't move deep enough to go to any clear solution. I get stuck and rather than showing my half scrubbed working I'd like to see how you'd do it from start to end with some thorough explanations. All 3 seem pretty similar, so working any one of them, whichever you find best, would be a big help.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thanks,

Wolf.

Mush
• December 19th 2008, 06:39 PM
Soroban
Hello, Lonehwolf!

Quote:

3) The graph of $f(x)\:=\:\frac{ax^2+bx+c}{x^2+qx+r}$ has the lines $x=1,\; x=3,\;y=2$ as asymptotes
and a turning point at $(0,1)$.

Find the constants $a, b, c, q,r$ and show that the graph has a second turning point.

Sketch the graph showing clearly its turning points
and its behaviour as it approaches the asymptotes.

Since $x=1\text{ and }x=3$ are vertical asymptoes,
. . the factors $(x-1)(x-3)$ appear in the denominator.

The function has the form: . $f(x) \:=\:\frac{ax^2 + bx + c}{x^2-4x+3}$
. . $\boxed{q = -4}\;\;\boxed{r = 3}$

Since $y = 2$ is a horizontal asymptote, $\lim_{x\to\infty} y \:=\:2$
. . This occurs when $\boxed{a = 2}$

The function has the form: . $f(x) \:=\:\frac{2x^2+bx + c}{x^2-4x+3}$

Since $(0,1)$ is on the graph: . $1 \:=\:\frac{0+0+c}{0-0+3} \quad\Rightarrow\quad \boxed{c \:=\:3}$

Differentiate: . $f\:\!'(x) \:=\:\frac{(4x+b)(x^2-4x+3) - (2x-4)(2x^2+bx+c)}{(x^2-4x+3)^2}$

There is a turning point at (0,1). .Hence: . $f\:\!'(0) = 0$

We have: . $f'(0) \:=\:\frac{(b)(3) - (-4)(c)}{3^2} \:=\:0\quad\Rightarrow\quad 3b + 4c \:=\:0 \quad\Rightarrow\quad b \:=\:-\frac{4}{3}c$

Since $c = 3\!:\;\;b \:=\:-\frac{4}{3}(3) \quad\Rightarrow\quad \boxed{b \:=\:-4}$

The function is: . $\boxed{f(x) \:=\:\frac{2x^2-4x+3}{x^2-4x+3}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For turning points, solve $f\:\!'(x) \:=\:0$

We have : . $f\:\!'(x) \:=\:\frac{(4x-4)(x^2-4x_3) - (2x-4)(2x^2-4x+3)}{(x^2-4x+3)^2} \:=\:0$

. . which simplifies to: . $\frac{-4x^2 + 6x}{(x^2-4x+3)^2} \:=\:0$

Then: . $-2x(2x-3) \:=\:0\quad\Rightarrow\quad x \:=\:0,\:\tfrac{3}{2}$

. . There is another turning point at: . $\left(\tfrac{3}{2},\:-2\right)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I believe the graph looks like this . . .
Code:

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• March 19th 2009, 03:09 PM
anaphase58
Hi, can someone help me sketch the curve y=x+cosx?

Also, I tried sketching the curve y = 2 (x)^1/2 + x and I'm thinking there's an asypmtote in it but I can't seem to prove that. Can someone also help me with this? please.(Worried)

( I'm supposed to show the intercepts , if any with the axes, any absolute or local extrema, the intervals over which f(x) is concave upward/downward, any points of inflection and the equations of any asypmtotes. )