Question

What are the coordinates of the turning point for the function f(x) = (x - 1)^3 - 3?

Answer 1

1) Don't submit a question with none of your own work showing.
2) Have you considered the 1st derivative?

Answer 2

Im sorry this is my first time ever asking a question.

Answer 3

No worries.

The best way to start out is the get the entire problem statement AND your best work in the VERY FIRST post.

Answer 4

I think the answer is -1,-3 because i learned in class that f(x) = (x-h)^3 +k = (h,k)

Answer 5

@amir_benjamin : Before we proceed, please describe what you think "turning point" means in this context. Can't do much until we agree on the meaning of terms.

Answer 6

The pont a which a polynomial function begins to turn?

Answer 7

So, you're saying you haven't tried the 1st derivative?

Turning Point is a at least a RELATIVE Minimum of Maximum. Of course, it might be global.

Answer 8

Some would say that an inflection point is also a "turning point."

Answer 9

Oh gosh, Im a bit confused what is the derivative?

Answer 10

You did not learn f(x) = (x-h)^3 +k = (h,k) in class, because that makes no sense at all.

Are you sure it's not f(x) = (x-h)^2 +k ==> (h,k)

Answer 11

Your right @tkhunny

Answer 12

My bad

Answer 13

Makes all the difference, doesn't it. :-)

Answer 14

It does, hah thank you

Answer 15

So: two questions for you:
How do you use the derivative to find turning points?
You are posting just part of the DEFINITION of the derivative.
If you're beyond learning that definition, then please use derivative rules to find the derivative of the given function.

Answer 16

I havent learned to much what a derivative is

Answer 17

\[f(x)=(x-1)^3-3\]

Answer 18

We don't care about the derivative in this class. We'll do that in a couple years. Right now, we just have a parabola.

Answer 19

Ok i know what a parabola is :D

Answer 20

You may have to improvise then. DRAW the given function. It should be obvious from your drawing where the "turning point" is. If you know what a parabola is, amir, then please get to work and graph the given function.

Answer 21

Oh gosh ok so i know my turning point is -1,-3 and i need to graph it right?

Answer 22

It would save you time if you can identify the vertex of this parabola and graph it, and then graph the rest of the parabola. You're being asked to graph the whole parabola, not just the turning point.

Answer 23

The vertex is the same as the turning point right?

Answer 24

Please, get to graphing, enough talk. In this case the turning point is the same as the vertex.

Answer 25

how do i put what I've drawn on here?

Answer 26

I have to take back my statement, "In this case the turning point is the same as the vertex." Why? Because this function has no vertex!

Use the Draw utility.

Answer 27

Blanket apology: two of us thought you were working with a parabola, when your function is actually a cubing function (which is an odd, not even, function).

Let's see your graph.

Answer 28

If it has no vertex how am i to graph it because i know if -1, -3 is our vertex then our axis of symmetry is x=-1 and additionally if i giveuse x=2 and x=4 i have y=(2-1)^3-3 which gives me (2,-2) and if i use x=4 y=(4-1)^3 -3 i have (4,24)

Answer 29

and i should just have to find the mirror points of (2,-2) and (4,24)

Answer 30

As I said, your function is a variation of the cubing function y=x^3. A rough graph of that follows: