anonymous
  • anonymous
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
Mathematics
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anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
I am confused as to what this question is even asking me.
anonymous
  • 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
  • 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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anonymous
  • anonymous
It notes that finding the turning point is similar to the vertex of a quadratic equation. Is this relevant?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i think so
anonymous
  • 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
  • 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
  • anonymous
Any ideas? I'm still very confused.
jdoe0001
  • jdoe0001
well.. if they're asking for "point of inflection", that's just a simple transformation
jdoe0001
  • jdoe0001
which I think may be what it's being asked
anonymous
  • 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
  • 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
  • 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
  • anonymous
That seems to fit the bill!
anonymous
  • anonymous
So what would be supposed "turning point," in which the substance changes matter?
anonymous
  • anonymous
anonymous
  • anonymous
Would it simply be the new origin point of the graph given the translations?

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