Question

How would I find the turning point here?

f(x) = (x - 4)^3 + 6

Answer 1

let me take a look

Answer 2

Thank you!

Answer 3

i cant help you with this one srry, i would of if i knew but i suck at science, only reason why i came here was because i though u look beautiful lol,

Answer 4

It's not science.. I just need to know how to find the turning point in a function.
Thanks, though.

Answer 5

I need to find the turning point of f(x) = (x - 4)^3 + 6, if anyone can help me out.

Answer 6

I'd think is when the function drops below the 0 for the y-axis

so that is, y = 0

Answer 7

Or would it be x=0?

Answer 8

well... is a temperature context... so if the T(x) gets high, the liquid evaporates

so I guess when the graph goes UP then the liquid temperature becomes too hot and turns to gas

either way, will be when y=0

Answer 9

well... dunno what "x" is ... I don't think is degrees

maybe it's ...

Answer 10

It's irrelevant what x and y are. The actual question just says to find the turning point of the function.

Answer 11

I mean, irrelevant if they stand for degrees or whatnot. I just need to know how to find the turning point of a function. .___. Isn't there some kind of formula?

Answer 12

well... I know what the function looks like... is not an even parent function, so it doesn't have a U-turn graph per se

Answer 13

To find the turning point of the function if it has one it will be where the derivative of the function is 0.

Answer 14

So basically, @whereismymind49 are you saying when f(0)?

Answer 15

hmmm.... that sounds more fit... yea... get the derivative... set it to 0, to get the tangent's value

Answer 16

No sorry. f'(x) = 0

Answer 17

Find that x point that makes f'(x) = 0

Answer 18

because the \(x^3\) parent function, doesn't really have a U-turn shape

Answer 19

f(0) = (0 - 4)^3 + 6
-4^3 + 6
-64 + 6 = 58

Is that the turning point?

Answer 20

?

Answer 21

No you need to do the derivative of the function first.

Answer 22

do you know what a derivative is?

Answer 23

Oh, how would I find the derivative?

Answer 24

hmmmm makes one wonder if you'r'e supposed to use the derivative in this exercise

Answer 25

I remember it faintly.. if either of you can kind of lead me on to find it I'm sure I could.

Answer 26

Though I honestly don't remember recently learning about it. :/

Answer 27

hmm well.. is calculus.... thus not sure if it'd apply in this case

Answer 28

It's algebra 2..

Answer 29

then I'm prone to think you're expected to find when the graph changes pattern... you can always just graph it in a calculator I gather

Answer 30

I'm not really allowed to use calculators. xD

Answer 31

ok... ahemm have you covered function transformations?

Answer 32

There needs to be SOME kind of formula. Because the question asks me to find the turning point and explain how I got there.

Answer 33

Yeah.

Answer 34

\(\large \textit{parent function}=x^3\qquad
\begin{array}{rllll}
f(x) = (x - 4)^3 &+ 6\\
\uparrow \quad &\uparrow \\
\textit{horizontal shift}&\textit{vertical shift}
\end{array}\)

Answer 35

so \(\bf x^3\) goes to the LEFT by 4 units, then UP by 6

I gather the turning point will be where it changes pattern, at y =0 for \(\bf x^3\)

Answer 36

woops . to the RIGHT, rather

Answer 37

so is a right h-shift and up v-shift

Answer 38

OHH

Answer 39

How did you know x^3 was the parent function?

Answer 40

beause \(\bf f(x) = (x - 4)^3 + 6\) is just a PADDED version of \(\bf x^3\)

Answer 41

Makes sense.. a little.

Answer 42

hehe, check your transformations chapter :)

it'll show the parent functions and then you pad them up with stuff, to "transform" the original

Answer 43

This could be a cubic function then?

Answer 44

it's a cubic one, yes, and parent odd function, they do not have a U-turn shape