Question

Please help with this graph !!!!

Answer 1

Note that 16=2*8, and that 8 is a perfect cube. Unfortunately, 5 is not a perfect cube.

Using this info, find the single x -value at which the denominator will be zero. This tells you the location of your vertical asymptote. I'd say it's OK to use a calculator to find this root.

Answer 2

It's nice that you know how to copy something and then annotate it.
However, your work would be better if you would label all of your notations. What did you intend to convey by (-1/(-16))?

Answer 3

How did you determine that the x-intercept is 2 something? My result was quite different.

Answer 4

Your y-intercept is correct.

Answer 5

@mathmale thank you. -1/-16 is the y intercept. I just plugged in 0 for the x. I got a huge number for the x intercept but I rounded to 2.07. I believe the vertical asymptote would be -2.07 as well?

Answer 6

By "huge number" you probably meant lots of digits after the decimal point.

To find the x-intercept(s), you must set the numerator = to 0 and solve for x.
Again, the answer is not 2.07 or close to that. Think: 2x^3 - 1 = 0, which means that:\[x^3=\frac{ 1 }{ 2 }\]

Answer 7

x=?

Answer 8

If you have a TI-83 or -84 calculator, you could find x as follows:

(1/2)^(1/3)

Answer 9

That's "the cube root of 1/2."

Answer 10

Sorry, it's been 8 minutes since I last heard from you. I need to get off the 'Net.

Answer 11

I'm sorry I was having dinner.
To find the x intercepts, I thought that I had to set the entire equation to 0. I did not know I had to set solely the numerator to 0.
For the other intercepts, do I plug in a value for x (into the entire equation) to find the y intercept?

Answer 12

lol

Answer 13

Am I doing at least one thing correctly?