MHF4U Reciprocal of a Linear Function by Roseanne L

Form:

zcmkoot — 2023-10-02 09:52:26 UTC

F(x)=1/kx-c

Restrictions (Excluded Values):

zcmkoot — 2023-10-02 09:55:57 UTC

To find the restrictions, set the denominator kx-c equal to zero and solve for x:
kx-c=0
kx=c
x=c/k
So, the function is undefined when x=c/k, which is the vertical asymptote.

Horizontal Asymptotes

zcmkoot — 2023-10-02 11:57:44 UTC

1. k>0➡️Left branch_negative, decreased slope.
Right branch_negative, increasing slope.
2. k<0➡️Left branch_positive, increasing slope.
Right branch_positive, decreased slope.

Equation For Horizontal Asymptotes

zcmkoot — 2023-10-02 11:58:55 UTC

y=0

X-Intercepts and Y-Intercepts

zcmkoot — 2023-10-02 12:00:55 UTC

To find the x-intercept, set f(x) equal to zero and solve for x: 1/kx-c=0
To find y-intercepts, set x=0
f(0)=1/-c

End Behaviour

zcmkoot — 2023-10-02 12:04:29 UTC

As x approaches positive or negative infinity, the function approaches zero because as x gets very large or very small, the impact of the constant c in the denominator becomes negligible compared to kx.

Concavity

zcmkoot — 2023-10-02 13:20:27 UTC

Example 1:

zcmkoot — 2023-10-02 13:31:34 UTC

Consider the reciprocal function f(x)=1/x−2

a) Find the vertical asymptote(s) of the function.

b) Determine if there is a horizontal asymptote, and if so, find it.

c) Calculate the x-intercept and y-intercept of the function.

Solution

zcmkoot — 2023-10-02 13:36:48 UTC

Consider the reciprocal function f(x)=1/x-2
a) To find the vertical asymptote(s), set the denominator (x−2) equal to zero and solve for x:
x−2=0
x=2
So, there is a vertical asymptote at x=2.

b) There is no horizontal asymptote for this function because the degree of the numerator (which is 1) is less than the degree of the denominator (which is also 1).

c) To find the x-intercept, set f(x)equal to zero and solve for

x:
1/x-2=0
This equation has no real solutions, meaning there are no x-intercepts.
To find the y-intercept, set x=0:
f(0)=1/0-2=-1/2
So, the y-intercept is at (0,-1/2)

Example 2:

zcmkoot — 2023-10-02 13:39:39 UTC

A population of bacteria in a petri dish is modeled by the function
P(t)=200/t+4, where P(t) represents the population (in thousands) at time t hours.

a) Find the vertical asymptote(s) of the population growth model.

b) Determine if there is a horizontal asymptote, and if so, find it. What does this asymptote represent in the context of the problem?

c) Calculate the time at which the population of bacteria is equal to 0.

Solution:

zcmkoot — 2023-10-02 13:42:45 UTC

a) To find the vertical asymptote(s), set the denominator (t+4) equal to zero and solve for t:
t+4=0
t=−4
So, there is a vertical asymptote at t=−4.

b) There is no horizontal asymptote for this population growth model because the degree of the numerator (which is a constant) is less than the degree of the denominator (which is 1).

c) To calculate the time at which the population of bacteria is equal to 0, set P(t) equal to zero and solve for t:
200/t+4=0
This equation has no real solutions, meaning the population never reaches zero.

In the context of the problem, the vertical asymptote at
t=−4 indicates that the population grows rapidly as time approaches t=−4, but it never actually reaches zero.