A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think so

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Any ideas? I'm still very confused.

  10. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    which I think may be what it's being asked

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That seems to fit the bill!

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jdoe0001

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.