Understanding graphing of rational functions

[arnuld...@gmail.com]

I have this equation: (4 - 3x - x^2)/(x - 3) and I am learning Pre-Calculus from coolmath.com and using https://www.desmos.com/calculator to graph.

I can easily calculate:

Y-intercept: f(0) = (0,-4/3)

X-intercepts: 4 - 3x - x^2 = 0 ==> (1,0) and (-4,0).

Vertical Asymptote: x - 3 = 0 => (3,0)

No Horizontal Asymptote

Oblique Asymptote: (4 - 3x - x^2)/(x - 3) ==> -x - 6

Graph came out to be like this: https://i.stack.imgur.com/OiAy9.png

Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)

Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.

-- arnuld
http://lispmachine.wordpress.com/
,= ,-_-. =.
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[Ross A. Finlayson]

Look at the order of terms,
notice it will have a term in x^1
and x^-1. It might not factor
and reduce but you can still
count the order of the terms,
as if you had.

See that this would establish
a chart that goes to to x and 1/x.

Then for the affine figure out cases
about scaling, translation, and rotation,
and flips here about establishing the origin
of the symmetry, i.e., the intersection
of the asymptotes.

There are still four quadrants,
they are just under affine
transformations.

Notice how the chart is much
like -x + 1/x, it still lives in the
same quadrants.

The coefficients are parameters.
Parameters are constants, here
the function is univariate, 'x' is
the variable.

The parameter's signs advise
what the resulting "skew
coordinate system" is.

This "skew coordinate system"
is that the quadrants (+,+) and (-,-)
are projected, that (+,-) and (-,+) has
that there is maintained an axis of
symmetry as f(x) = x and f(x) = -x,
besides the usual f(x) = 0 and x = 0.

Then where there are particular
critical values (but the above is
about rational functions) that
give different shapes about
them, those are called shape
parameters. You can find these
for example in theories of
distributions (and what
parameters effect different
shapes of the distributions).

[FredJeffries]

On Saturday, July 21, 2018 at 8:10:48 AM UTC-7, arnuld...@gmail.com wrote:

> I have this equation: (4 - 3x - x^2)/(x - 3) and I am learning Pre-Calculus from coolmath.com and using https://www.desmos.com/calculator to graph.
>
> I can easily calculate:
>
> Y-intercept: f(0) = (0,-4/3)
>
> X-intercepts: 4 - 3x - x^2 = 0 ==> (1,0) and (-4,0).
>
> Vertical Asymptote: x - 3 = 0 => (3,0)
>
> No Horizontal Asymptote
>
> Oblique Asymptote: (4 - 3x - x^2)/(x - 3) ==> -x - 6
>
> Graph came out to be like this: https://i.stack.imgur.com/OiAy9.png
>
> Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)

Why do you think there should be three parts? You've said that there's only one vertical asymptote which divides the x-axis into two sections.

What is the "middle" of which you speak?

> Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.

aHA!! That's why this is called "PRE-calculus". You will learn the tricks to find maxima and minima and the direction of concavity and inflection points and ... when you get to Calculus (Yes, I promise).

Thank you for asking these questions. It had been a long time since I had looked at any of this.

[Barry Schwarz]

On Sat, 21 Jul 2018 08:10:41 -0700 (PDT), arnuld...@gmail.com
wrote:

>I have this equation: (4 - 3x - x^2)/(x - 3) and I am learning Pre-Calculus from coolmath.com and using https://www.desmos.com/calculator to graph.

This is not an equation. It is just an expression. The equation you
are graphing is y = (4 - 3x - x^2)/(x - 3).

>I can easily calculate:
>
>Y-intercept: f(0) = (0,-4/3)
>
>X-intercepts: 4 - 3x - x^2 = 0 ==> (1,0) and (-4,0).
>
>Vertical Asymptote: x - 3 = 0 => (3,0)
>
>No Horizontal Asymptote
>
>Oblique Asymptote: (4 - 3x - x^2)/(x - 3) ==> -x - 6
>
>Graph came out to be like this: https://i.stack.imgur.com/OiAy9.png
>
>Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)

To explain why, consider trying to find the value of x for which y is
10 (a convenient point in the middle). This quickly leads to a
quadratic in x with a negative discrimant. This tells you that the
value of x is complex rather than real. So it's not that there is
nothing in the middle. It's that real values of x do not produce such
a middle.

>Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.

Consider the common hyperbola xy = k or the equivalent xy - k = 0. You
already know what this looks like. Replacing x and y with x-x0 and
y-y0 translates the curve away from the origin but doesn't change its
shape in any way. But you can multiply it out to obtain terms in xy,
x, y, and a constant.

Now if you take the equation above, multiply by the denominator, and
combine terms, you also end up with an equation containing terms in
xy, x, y, and constant. This implies that the graph looks like a
hyperbola but probably rotated and translated.

As far as the asymptotes, examine the limits of y' as x increases (or
decreases) without bound. Inflection points occur where y' is 0.

--
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[Barry Schwarz]

On Sun, 22 Jul 2018 18:06:38 -0700, Barry Schwarz <schw...@dqel.com>
wrote:

>On Sat, 21 Jul 2018 08:10:41 -0700 (PDT), arnuld...@gmail.com
>wrote:
>
>>I have this equation: (4 - 3x - x^2)/(x - 3) and I am learning Pre-Calculus from coolmath.com and using https://www.desmos.com/calculator to graph.
>
>This is not an equation. It is just an expression. The equation you
>are graphing is y = (4 - 3x - x^2)/(x - 3).
>
>>I can easily calculate:
>>
>>Y-intercept: f(0) = (0,-4/3)
>>
>>X-intercepts: 4 - 3x - x^2 = 0 ==> (1,0) and (-4,0).
>>
>>Vertical Asymptote: x - 3 = 0 => (3,0)
>>
>>No Horizontal Asymptote
>>
>>Oblique Asymptote: (4 - 3x - x^2)/(x - 3) ==> -x - 6
>>
>>Graph came out to be like this: https://i.stack.imgur.com/OiAy9.png
>>
>>Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)
>
>To explain why, consider trying to find the value of x for which y is
>10 (a convenient point in the middle). This quickly leads to a

Obviously -10.