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

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

yes

- anonymous

i think so

- anonymous

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.

- jdoe0001

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

- anonymous

Any ideas? I'm still very confused.

- jdoe0001

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

- jdoe0001

which I think may be what it's being asked

- anonymous

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...

- anonymous

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.

- jdoe0001

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 \(x^3\)

- anonymous

That seems to fit the bill!

- anonymous

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

- anonymous

- anonymous

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

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