Producto Integrador Global Matematicas 3 UANL

[pic 1]Producto Integrador Global

Mathematics 3: Analytic Geometry

Prof: Sergio Bustamante

Student: Miguel Angel Garcia Cruz

#13

1746113

Group: 301-B

17/11/15

Introduction

In this final project of math 3 I will represent all the stages, this means that I will show almost every type of problem that we saw in this semester, this 8 problems will include pictures and explanations for the good of the reader.

This problems will be based on the next topics:

1. Linear functions
2. Quadratic functions
3. Derivates of superior order
4. Rational functions
5. Rational functions of second degree
6. Irrational function
7. Variation function
8. Logarithms
9. Exponential in various contexts

Exercises

Linear function

A linear function has the following form

y = f(x) = a + bx

A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y.

a is the constant term or the y intercept. It is the value of the dependent variable when x = 0.

b is the coefficient of the independent variable. It is also known as the slope and gives the rate of change of the dependent variable

Graphing a linear function

To graph a linear function:

1. Find 2 points which satisfy the equation

2. Plot them

3. Connect the points with a straight line

Example:

[pic 2]y = 25 + 5x

let x = 1

then

y = 25 + 5(1) = 30

let x = 3

then

y = 25 + 5(3) = 40

Quadratic Function

Find the x-intercepts and vertex of y = –x2 – 4x + 2.

Since it is so simple to find the y-intercept (and it will probably be a point in my T-chart anyway), they are only asking for the x-intercepts this time. To find the x-intercept, I set y equal 0 and solve:

0 = –x2 – 4x + 2

x2 + 4x – 2 = 0

[pic 3]

For graphing purposes, the intercepts are at about (–4.4, 0) and (0.4, 0). (When I write down the answer, I will of course use the "exact" form, with the square roots; my calculator's decimal approximations are just for helping me graph.)

To find the vertex, I look at the coefficients: a = –1 and b = –4. Then:

h = –(–4)/2(–1) = –2

To find k, I plug h = –2 in for x in y = –x2 – 4x + 2, and simplify:

k = –(–2)2 – 4(–2) + 2 = –4 + 8 + 2 = 10 – 4 = 6

Now I'll find some additional plot points, to help me fill in my graph:

T-chart

[pic 4]Note that I picked x-values that were centered around the x-coordinate of the vertex. Now I'll plot the parabola:

graph of y = -x^2 - 4x + 2  

Derivates of superior order

A higher order derivative is known as the second derivative of the function, that is, if f (x) is a function exists and its first derivative f '(x).[pic 5]

similarly can obtain higher order derivatives, but it is necessary to clarify that the derivatives of a function depend on the characteristics of the function and possible, and it often happens that some derivatives exist but not all areas despite which can be calculated with formulas.    

Rational Functions

Graph the following: [pic 6]

First I'll find the vertical asymptotes, if any, for this rational function. Since I can't graph where the function doesn't exist, and since the function won't exist where there would be a zero in the denominator, I'll set the denominator equal to zero to find any forbidden points:

[pic 7]x – 1 = 0 
x = 1

So I can't have x = 1, and therefore I have a vertical asymptote there.

I'll dash this in on my graph:

[pic 8]Next I'll find the horizontal or slant asymptote. Since the numerator and denominator have the same degree (they're both linear), the asymptote will be horizontal, not slant, and the horizontal asymptote will be the result of dividing the leading coefficients:

y = 2/1 = 2

                                                        I'll dash this in, too:

           Next, I'll find any x- or y-intercepts.

[pic 9]x = 0:  y = (0 + 5)/(0 – 1) = 5/–1 = –5 
y = 0:  0 = (2x + 5)/(x – 1) 
              0 = 2x + 5 
        –5 = 2x 
     –2.5 = x

Then the intercepts are at (0, –5)
and (–2.5, 0). I'll sketch these in:    

Now I'll pick a few more x-values, compute the corresponding y-values, and plot a few more points.   Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

[pic 10]I mostly picked x-values near the middle of the graph: because of the horizontal asymptote, I already have a good idea of what the graph does off to the sides. (It can be a good idea to do a point or two near the ends anyway, as a check on your work.) Also, since I had no intercepts on the right-hand side of the vertical asymptote to give me hints as to what was happening with the graph, I needed more points there to show me what was going on.

...