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Anonymous Fri 04 Dec 18:18:34 2020 No.12419794 View Reply Original Report

Quoted By: >>12419812 >>12419818 >>12419941 >>12419967 >>12420013 >>12421595 >>12421648

How do you find the x intersects with this function again?
Obviously you gotta set the equation at 0 but i don't know how to solve it.

Anonymous

Anonymous Fri 04 Dec 2020 18:22:47 No.12419805 Report

Quoted By: >>12419813

Complete the square

Anonymous

Anonymous Fri 04 Dec 2020 18:24:40 No.12419811 Report

Quoted By: >>12419817

when x² - 4 = 0 then it intersects the x-axis.

Is that what you need?

Anonymous

Anonymous Fri 04 Dec 2020 18:24:43 No.12419812 Report

Quoted By:

>>12419794
Factor the top. it has vertical asymptotes at 0 and 6. Factoring the top tells you it has 0s at 2 and -2.

Anonymous

Anonymous Fri 04 Dec 2020 18:25:07 No.12419813 Report

Quoted By:

>>12419805
Oh wow, i'm actually impulsive

Anonymous

Anonymous Fri 04 Dec 2020 18:27:09 No.12419817 Report

Quoted By:

>>12419811
shit I meant when

x^2 - 4=0

x^2 = 4

Anonymous

Anonymous Fri 04 Dec 2020 18:27:11 No.12419818 Report

Quoted By:

>>12419794
You have to be at least 18 to post on 4chan

Anonymous

Anonymous Fri 04 Dec 2020 18:38:47 No.12419855 Report

Quoted By:

distinction of cases, possibly calculate limit values for completeness maybe?
in any case, consider the numerator and denominator individually

Anonymous

Anonymous Fri 04 Dec 2020 19:04:28 No.12419932 Report

Quoted By: >>12421591 >>12421637

f(x) = (x^2 - 4) / (x^2 - 6x)

Finding f(x) = 0:
(x^2 - 4) / (x^2 - 6x) = 0
x^2 - 4 = 0 | (x^2 - 6x) != 0
x^2 = 4
x = +- sqrt(4)
x = +-2

Finding vertical asymptotes:
f(x) approaches infinity
(x^2 - 4) / (x^2 - 6x) -> inf
x^2 - 6x = 0 | x^2 - 4 != 0
x(x - 6) = 0
x = 0 v x = 6

Finding horizontal asymptotes:
x approaches infinity while f(x) is finite.
(x^2 - 4) / (x^2 - 6x) | x-> inf
We divide both the upper and lower term with x^2 to simplify.
(1 - 4/x^2) / (1 - 6/x) | x -> inf
Anything divided by inf is 0.
(1 - 0) / (1 - 0)
1/1
1
As x approaches inf, f(x) approaches 1. The vertical asymptote is at f(x) = 1

Feel free to ask if anything is unclear.

Anonymous

Anonymous Fri 04 Dec 2020 19:09:23 No.12419941 Report

Quoted By:

>>12419794
Back to school kid

Anonymous

Anonymous Fri 04 Dec 2020 19:22:31 No.12419967 Report

Quoted By:

>>12419794
cancel the x2 and then do your elementary school math

Anonymous

Anonymous Fri 04 Dec 2020 19:43:33 No.12420013 Report

Quoted By:

>>12419794
Its +2 and -2, because the denominator cant be 0, so the fraction is 0 when the numerator is 0, you can factor the numerator since its equal to (×+2)(x-2).

Anonymous

Anonymous Sat 05 Dec 2020 05:50:37 No.12421591 Report

Quoted By:

>>12419932
>(x^2 - 6x) != 0
(x^2 - 6x) ?0?!

Anonymous

Anonymous Sat 05 Dec 2020 05:52:28 No.12421595 Report

Quoted By:

>>12419794
Well it's equal to zero when the top is zero. So work that out.

Anonymous

Anonymous Sat 05 Dec 2020 06:09:19 No.12421637 Report

Quoted By:

>>12419932
-> inf == *inf?

Anonymous

Anonymous Sat 05 Dec 2020 06:13:41 No.12421648 Report

Quoted By:

>>12419794
x intercepts the y axis when y = 0.
f(x) = y = 0
x^2 = 4
x=+/-2

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