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anonymous
 one year ago
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
anonymous
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.0I am confused as to what this question is even asking me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do 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 (x4) (x4) (x4) + 6 we have to multiply that out to get it in the form so we can find a, b and c first do (x4)(x4) (x^2 8x+16)(x4) 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
 one year ago
Best ResponseYou've already chosen the best response.0and 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It notes that finding the turning point is similar to the vertex of a quadratic equation. Is this relevant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1hmmm 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 Uturn as even ones do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Any ideas? I'm still very confused.

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well.. if they're asking for "point of inflection", that's just a simple transformation

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1which I think may be what it's being asked

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1right... the socalled "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
 one year ago
Best ResponseYou've already chosen the best response.0That seems to fit the bill!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what would be supposed "turning point," in which the substance changes matter?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it simply be the new origin point of the graph given the translations?
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