A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

please help me..... The curve whose equation is y=ax^3 + bx^2 + cx + d has a point of inflexion at (-1,4), a turning point when x=2, and it passes through the point (3,-7). Find the values of a, b, c and d.

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

    a=3/11 b=9/11 c=-72/11 d=-23/11 \[y'=3ax ^{2}+2bx+c\] \[y''=6ax+2b\] inflection poit at (-1,4) means the second derivative =0 at x=1 which gives \[0=-6a+2b\] which means b=3a a turning point at x=2 means the first derivative =0 at x=2 which gives\[12a+4b+c=0\] So 24a+c=0 and c=-24a use the inf that f(3)=-7 and f(-1)=4 to solve for a and d

  2. 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...


  • 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.