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1) Don't submit a question with none of your own work showing. 2) Have you considered the 1st derivative?
Im sorry this is my first time ever asking a question.
No worries. The best way to start out is the get the entire problem statement AND your best work in the VERY FIRST post.
I think the answer is -1,-3 because i learned in class that f(x) = (x-h)^3 +k = (h,k)
@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.
The pont a which a polynomial function begins to turn?
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.
Some would say that an inflection point is also a "turning point."
Oh gosh, Im a bit confused what is the derivative?
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)
Your right @tkhunny
Makes all the difference, doesn't it. :-)
It does, hah thank you
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.
I havent learned to much what a derivative is
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.
Ok i know what a parabola is :D
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.
Oh gosh ok so i know my turning point is -1,-3 and i need to graph it right?
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.
The vertex is the same as the turning point right?
Please, get to graphing, enough talk. In this case the turning point is the same as the vertex.
how do i put what I've drawn on here?
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.
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.
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)
and i should just have to find the mirror points of (2,-2) and (4,24)
As I said, your function is a variation of the cubing function y=x^3. A rough graph of that follows:|dw:1449970061856:dw|
no not really
unless we count the slight turn towards the right by the origin
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.
Bye thanks for the help
Im still a bit confused
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.