Taylor Blog #13

This week we covered two major topics
these topic included:
Horizontal asymptotes
and
curve sketching

To find a horizontal asymptote there are three rules to follow
these rules apply to the degree of the leading coefficients of the polynomials in both the numerator and denominator
These rules are:

* If the top degree of the leading coefficient is equal to the bottom degree of the leading coefficient then you take the coefficient of the highest degree in the numerator and put it over the coefficient of the highest degree in the denominator

Example:
What is the horizontal asymptote of
2x^2+5 / 3x^2+1

because the degrees of the leading coefficients are equivalent there is a horizontal asymptote at 2/3


*If the top degree of the leading coefficient is greater than bottom degree of the leading coefficient then there is a horizontal asymptote at - infinity , infinity. In other words there is no horizontal asymptote.


Example:
What is the horizontal asymptote of
2x^2+5 / 3x+1

because the degree of the leading coefficient in the top is greater than the leading coefficient in the bottom the result is
-infinity, infinity and therefore there are no horizontal asymptotes.




*If the top degree of the leading coefficient is less than bottom degree of the leading coefficient then there is a horizontal asymptote at 0.


Example:
What is the horizontal asymptote of
2x+5 / 3x^2+1

because the degree of the leading coefficient in the top is less than the leading coefficient in the bottom the horizontal asymptote is 0.




For curve sketching problems there are five steps to follow

these steps are:

#1
find the domain and find the range
The domain is found by setting the bottom equal to zero, solving for x, and setting up intervals
The range is found by finding the horizontal asymptotes and setting up intervals

#2
Find the x intercepts, y intercepts, vertical asymptotes, and horizontal asymptotes.
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
Vertical asymptotes and horizontal asymptotes have already been found at this point
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals

#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
Solve for x
check differentiability
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative and solve
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))

#4
Do second derivative test
take the derivative of the first derivative
set it equal to zero
solve for x
check differentiability
set up intervals (-infinity, pt)U(pt,infinity)
plug into the second derivative and solve
((if you get a positive number the function is concave up, If you get a negative number the function is concave down))
((any point on the interval where a number switches from concave up to concave down or vise versa there is a point of inflection))

#5
Plot all important information on a sketched graph.

IMPORTANT INFORMATION INCLUDES:
domain intervals
range intervals
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
whether the intervals of the first derivative test are increasing or decreasing
where there is a max or a min
whether the intervals of the second derivative test are concave up or concave down



Example:
y= 2(x^2-9)/x^2-4


#1
find the domain and find the range

The domain:
x^2-4=0
x^2=4
x = +/- 2
(-infinity,-2)(-2,2)(2,infinity)

The range is found by finding the horizontal asymptotes and setting up intervals
y= 2
(-infinity,2)(2,infinity)

#2

X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
2(x^2-9)=0 therefore: x=+/-3
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
2(0^2-9)/0^2-4
2(-9)/-4
-18/-4
9/2
(0,9/2)

your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals

#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
x^2-4(4x)-[2x^2-18(2x)]/(2^2-4)^2
20x/(x^2-4)^2
Solve for x
20x=0
x=0
check differentiability
vertical asymptotes at +/- 2
Set up intervals (-infinity, pt)U(pt, infinity)
(-infinity,-2)(-2,0)(0,2)(2,infinity)
Plug in value on the interval into the derivative and solve
f(-3) negative therefore decreasing
f(-1) negative therefore decreasing
f(1) positive therefore increasing
f(3) positive therefore increasing

MIN @ x=0
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))

#4
Do second derivative test
take the derivative of the first derivative
-20(3x^2+4)/(x^2-4)^3

set it equal to zero
-20(3x^2+14)=0
solve for x
x=sqrt -4/3 therefore does not apply

check differentiability
vertical asymptotes at +/-2
set up intervals (-infinity, pt)U(pt,infinity)
(-infinity, -2)(-2,2)(2, infinity)
plug into the second derivative and solve
f(-3) negative concave down
f(0) positive concave up
f(3) negative concave down


Then you would proceed to plot all important information on a sketched graph