## 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

1. anonymous

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

2. 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

3. 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

4. anonymous

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

5. anonymous

yes

6. anonymous

i think so

7. 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.

8. 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

9. anonymous

Any ideas? I'm still very confused.

10. jdoe0001

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

11. jdoe0001

which I think may be what it's being asked

12. 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...

13. 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.

14. 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 $$x^3$$

15. anonymous

That seems to fit the bill!

16. anonymous

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

17. anonymous

@jdoe0001

18. anonymous

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