Turn on thread page Beta
    • Thread Starter
    Offline

    12
    ReputationRep:
    Question: The curve y=f(x) passes through the origin.
    Given that \frac{dy}{dx} = 3x^{2} - 8x -5, find the coordinates of the other points where the curve crosses the x-axis.

    So i started integrating to get y=x^{3}-4x^{2}- 5x+c. Since the curve goes through origin, i assume that c=0. how do i find the co-ordinates of the other points where the curve crosses the x-axis.

    Should i just just substitute random number like 1 or 2 for x and work out what their y co-ordinate is?
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by Chelsea12345)
    Question: The curve y=f(x) passes through the origin.
    Given that \frac{dy}{dx} = 3x^{2} - 8x -5, find the coordinates of the other points where the curve crosses the x-axis.

    So i started integrating to get y=x^{3}-4x^{2}- 5x+c. Since the curve goes through origin, i assume that c=0. how do i find the co-ordinates of the other points where the curve crosses the x-axis.

    Should i just just substitute random number like 1 or 2 for x and work out what their y co-ordinate is?
    You're trying to solve x^3-4x^2-5x=0. Try factorizing that. There is actually no need to use the factor theorem.

    hint
    If you were to have x^5-25x=0 you could say that x(x^4-25)=0 \Rightarrow x=0, \pm \sqrt{5}.
    Offline

    11
    ReputationRep:
    x(x-5)(x+1)
    So it crosses at (0,0) (5,0) (-1,0)
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Kvothe the Arcane)
    You're trying to solve x^3-4x^2-5x=0. Try factorizing that. There is actually no need to use the factor theorem.

    hint
    If you were to have x^5-25x=0 you could say that x(x^4-25)=0 \Rightarrow x=0, \pm \sqrt{5}.
    Thankyou for the hint btw! so i took out the x to x(x^2 -4x -5)=0.
    Then i added 5 to the other side to get x^2 - 4x = 5. What should i do now to solve this?
    • Community Assistant
    • Very Important Poster
    Offline

    20
    ReputationRep:
    Community Assistant
    Very Important Poster
    (Original post by Chelsea12345)
    Thankyou for the hint btw! so i took out the x to x(x^2 -4x -5)=0.
    Then i added 5 to the other side to get x^2 - 4x = 5. What should i do now to solve this?
    You are welcome.
    Do you know how you'd factorize the quadratic equation x^2-4x-5=0?
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by Kvothe the Arcane)
    You are welcome.
    Do you know how you'd factorize the quadratic equation x^2-4x-5=0?
    Yes i didnt realise it was a quadratic inside the brackets!
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Mr Moon Man)
    x(x-5)(x+1)
    So it crosses at (0,0) (5,0) (-1,0)
    Not cool. Please read the Guide To Posting at the top of the Maths Forum and pay close attention to the bit about not providing full solutions.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 8, 2017
Poll
Favourite type of bread
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.