anonymous
  • anonymous
What are the coordinates of the turning point for the function f(x) = (x - 1)^3 - 3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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tkhunny
  • tkhunny
1) Don't submit a question with none of your own work showing. 2) Have you considered the 1st derivative?
anonymous
  • anonymous
Im sorry this is my first time ever asking a question.
tkhunny
  • tkhunny
No worries. The best way to start out is the get the entire problem statement AND your best work in the VERY FIRST post.

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More answers

anonymous
  • anonymous
I think the answer is -1,-3 because i learned in class that f(x) = (x-h)^3 +k = (h,k)
mathmale
  • mathmale
@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.
anonymous
  • anonymous
The pont a which a polynomial function begins to turn?
tkhunny
  • tkhunny
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.
mathmale
  • mathmale
Some would say that an inflection point is also a "turning point."
anonymous
  • anonymous
Oh gosh, Im a bit confused what is the derivative?
tkhunny
  • tkhunny
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)
anonymous
  • anonymous
Your right @tkhunny
anonymous
  • anonymous
My bad
tkhunny
  • tkhunny
Makes all the difference, doesn't it. :-)
anonymous
  • anonymous
It does, hah thank you
mathmale
  • mathmale
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.
anonymous
  • anonymous
I havent learned to much what a derivative is
mathmale
  • mathmale
\[f(x)=(x-1)^3-3\]
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
Ok i know what a parabola is :D
mathmale
  • mathmale
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.
anonymous
  • anonymous
Oh gosh ok so i know my turning point is -1,-3 and i need to graph it right?
mathmale
  • mathmale
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.
anonymous
  • anonymous
The vertex is the same as the turning point right?
mathmale
  • mathmale
Please, get to graphing, enough talk. In this case the turning point is the same as the vertex.
anonymous
  • anonymous
how do i put what I've drawn on here?
mathmale
  • mathmale
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.
mathmale
  • mathmale
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.
anonymous
  • anonymous
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)
anonymous
  • anonymous
and i should just have to find the mirror points of (2,-2) and (4,24)
mathmale
  • mathmale
As I said, your function is a variation of the cubing function y=x^3. A rough graph of that follows:|dw:1449970061856:dw|
anonymous
  • anonymous
no not really
anonymous
  • anonymous
unless we count the slight turn towards the right by the origin
mathmale
  • mathmale
The graph of your function does not go through the origin, as the graph of y=x^3 does. There is a "turning point" at (1,-3). Graph (1,-3) and then try sketching the whole graph. Its shape is the same as the one I've drawn for y ou. I will be absent from OpenStudy for about 30 minutes. Look up "turning point." What does it mean here? Not every function that has a turning point has a vertex and/or is a parabola. Later.
anonymous
  • anonymous
Bye thanks for the help
anonymous
  • anonymous
Im still a bit confused
Owlcoffee
  • Owlcoffee
The turning points are also the local extreme points of the function, meaning that you'd have to find the derivative first and then find those values of "x" that make the derivative zero.

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