Question

In your lab, a substance's temperature has been observed to follow the function T(x) = (x − 4)3 + 6. The turning point of the graph is where the substance changes from a liquid to a gas. Using complete sentences in your written answer, explain to your fellow scientists how to find the turning point of this function

Answer 1

I am confused as to what this question is even asking me.

Answer 2

Do you know about the turning point formula?

one year agoReport Abuse
1 attachments
turning_point.png turning_point.png
I think the a,b, and c are from
a x^3 + b x^2 +c x + d

unfortunately, you have (x - 4)^3 + 6
which is (x-4) (x-4) (x-4) + 6
we have to multiply that out to get it in the form so we can find a, b and c

first do (x-4)(x-4)
(x^2 -8x+16)(x-4)
that means you function is
(x−4)3+6=x3−12x2+48x−64+6=x3−12x2+48x−58

the formula says -B in front. B is -12, so you have - -12 or +12

Answer 3

and you are adding the sqr(0)
so it is
12±0√3

sqrt(0) is 0 so that is
the answer is x=12/3 = 4
that is the turning point of this curve
Here is a graph
one year agoReport Abuse
1 attachments
graph.png graph.png

Answer 4

It notes that finding the turning point is similar to the vertex of a quadratic equation. Is this relevant?

Answer 5

yes

Answer 6

i think so

Answer 7

I am still confused. Are you sure this is the right method? I'm just unsure as to what exactly the turning point is on a graph.

Answer 8

hmmm right, I wonder that myself, since it's not an even exponent or "even" equation, is a cubic, or "odd" one
and "odd" ones do not have a "vertex" or U-turn as even ones do

Answer 9

Any ideas? I'm still very confused.

Answer 10

well.. if they're asking for "point of inflection", that's just a simple transformation

Answer 11

which I think may be what it's being asked

Answer 12

I mean it must be something about how they hinted to the fact that it similar to finding the vertex of a quadratic. Is it maybe the point at which a cubic graph begins going up or down? But I guess that would just be the point of origin in most cases...

Answer 13

I hope you know what I mean. Like the point from which the left arm, in this case the down arm, ends, and the right arm, in this case the up arm, begins.

Answer 14

right... the so-called "stationary point of inflection"
if that's what you're being asked... then
\(\textit{function transformations}
\\ \quad \\

\begin{array}{llll}

\begin{array}{llll} shrink\ or\\ expand\\ by\ {\color{purple}{ A}}\end{array}
\qquad
\begin{array}{llll} vertical\\ shift\\ by \ {\color{green}{ D}} \end{array}

\begin{array}{llll}{\color{green}{ D}} > 0& Upwards\\ {\color{green}{ D}} < 0 & Downwards\end{array}
\\

% template start
\qquad \downarrow\qquad\qquad\quad\ \downarrow\\
T(x) = {\color{purple}{ A}} ( x {\color{red}{ -4}} ) + {\color{green}{ 6}}\\
%template end
\qquad\qquad \quad \uparrow \\

\qquad\begin{array}{llll} horizontal\\ shift\\ by \ {\color{red}{ C}}\end{array}

\begin{array}{llll}{\color{red}{ C}} > 0 & to\ the\ left\\ {\color{red}{ C}} < 0& to\ the\ right\end{array}
\end{array}\)


so.. notice the function transformation
keeping in mind that the "parent function" is \(x^3\)

Answer 15

That seems to fit the bill!

Answer 16

So what would be supposed "turning point," in which the substance changes matter?

Answer 17

@jdoe0001

Answer 18

Would it simply be the new origin point of the graph given the translations?