How do you graph f(x)=2x−1/x^3−9x using holes, vertical and horizontal asymptotes, x and y intercepts?

tuzkutimonq4

tuzkutimonq4

Answered question

2022-09-11

How do you graph f ( x ) = 2 x - 1 x 3 - 9 x using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer & Explanation

Kiera Moreno

Kiera Moreno

Beginner2022-09-12Added 9 answers

FIRST thing we ALWAYS do is making the function easier:

f ( x ) = 2 x - 1 ( x 3 - 9 x )
2 x - 1 x ( x 2 - 9 )
2 x - 1 x ( x 2 - 9 )
2 x - 1 x ( x - 3 ) ( x + 3 ) because ( a 2 - b 2 ) = ( a - b ) ( a + b )

Now we can "easily" see that x MUST NOT be 0,3,-3 because we know that we must not divided by 0.
We can also see that (2x−1=0) if ( x = 1 2 )

vertical asymptotes:

Therefore we now know that the function has 3 vertical asymptotes (we get them when x=0 in the "down side of the equation"):
x=0
x=3
x=−3

horizontal asymptotes:
(I don't know if you know how to use lim so I will explain in an alternative way)

horizontal asymptotes is a question of a kind of "What does the function "act" when x is REALLY HUGE (and/or negative HUGE)?

To solve it easily, lets check with a calculator what happens when x comes is really big, for example I will check the positives:
f 1 000 000 = 1.999999 10 - 12 = 0.0000000000000001999
f 1 00 000 000 = 1.9999 10 - 15 = 0.0000000000000000001999

So, the function goes to y=0 for very big x-es.
On the same way, we can chek the very negative x-es:
f 1 000 000 = 2.000001 10 - 12 = 0.00000000000000020001

So, in that case, the function goes to y=0 for very negative big x-es.

therefore, the horizontal asymptote is:
y=0

x and y intercepts:
Now that we have an "easier" function:
( f ( x ) = 2 x - 1 x ( x - 3 ) ( x + 3 ) )

We check both "x=0" and y=0":
f ( 0 ) = 2 0 - 1 0 ( 0 - 3 ) ( 0 + 3 ) = - 1 0 = N a N
But we already knew it because x=0 is one of ours' vertical asymptotes...
Now lets check what if f ( x ) = 0 what is x?
2 x - 1 x ( x - 3 ) ( x + 3 ) = 0
( 2 x - 1 ) = 0 (for x in not 1,3,-3)
2 x = 1
x = 1 2

So we have only one point A = ( 1 2 , 0 )

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