C1 Sketching Quadratic solution, where does the bottom line come from?

Announcements Posted on
    • Thread Starter
    Offline

    3
    ReputationRep:
    We haven't been taught how to do it, but it is for our homework. Ihave gone on the solution bank to try and work out how to do it, and this is the answer and solution (image attached).

    I can follow and understand it up to where it says the x axis crossing points are (-1,0) and (-4,0) because they are the values of x when you solve it via factorising.
    However, I cannot work out for the life of me where the whole bottom line comes from, and what it means when it says: x=0, y=4 so y axis crossing point is (0,4). Can you please explain where the x=0 and y=4 comes from please?

    As I said, I haven't been taught how to do it (we have to teach ourselves this bit for homework) and I can't understand it.
    Thanks!
    Attached Images
     
    Online

    3
    ReputationRep:
    it means that when x =0, y=4.
    Offline

    2
    ReputationRep:
    Well sub x=0 into y=x^2+5x+4.
    Offline

    2
    ReputationRep:
    The point at which the graph crosses the y-axis is the point at which  x = 0 . If you substitute  x = 0 into  x^2 + 5x + 4 , the answer is 4.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by an_atheist)
    Well sub x=0 into y=x^2+5x+4.
    When solving a question like this though, would you always do it exactly the same way as this solution? By that, I mean would you work out both x values by factorising (like they did in this solution), and then always sub x=0 into work out the y-axis point?Like do you always sub y=0 when you work out the x axis value, and then you also sub x=0 in to work out the y coordinate?
    Offline

    2
    ReputationRep:
    (Original post by blobbybill)
    When solving a question like this though, would you always do it exactly the same way as this solution? By that, I mean would you work out both x values by factorising (like they did in this solution), and then always sub x=0 into work out the y-axis point?Like do you always sub y=0 when you work out the x axis value, and then you also sub x=0 in to work out the y coordinate?
    Essentially, yes. Note though that if you substitute  x = 0 into a quadratic  y = ax^2 + bx + c the y coordinate will always be  c
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Tau)
    Essentially, yes. Note though that if you substitute  x = 0 into a quadratic  y = ax^2 + bx + c the y coordinate will always be  c
    Thanks. I have forgotten what this highlighted bit comes from, what it means and why you do it. I get that you sub y=0 to work out the x coordinates, but how do you know what you factorise into in order to get (x+1)(x+4)? What equation does that derive from?

    ie, how do you know what to factorise, in order to get (x+1)(x+4)??. Sorry, I have just gotten really confused.


    Name:  Screenshot_7.png
Views: 14
Size:  18.6 KB
    Offline

    2
    ReputationRep:
    (Original post by blobbybill)
    Thanks. I have forgotten what this highlighted bit comes from, what it means and why you do it. I get that you sub y=0 to work out the x coordinates, but how do you know what you factorise into in order to get (x+1)(x+4)? What equation does that derive from?

    ie, how do you know what to factorise, in order to get (x+1)(x+4)??. Sorry, I have just gotten really confused.
    You have an equation of the form  y = ax^2 + bx +c or in your case,  y = x^2 + 5x + 4 . If  y = 0 then  x^2 + 5x +4 = 0 , so  (x+1)(x+4) = 0 .
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Tau)
    You have an equation of the form  y = ax^2 + bx +c or in your case,  y = x^2 + 5x + 4 . If  y = 0 then  x^2 + 5x +4 = 0 , so  (x+1)(x+4) = 0 .
    So to find the x coordinates you sub y=0 into it and factorise it to get the values of x, and to find the y coordinate you sub x=0 into the equation?
    Offline

    2
    ReputationRep:
    (Original post by blobbybill)
    So to find the x coordinates you sub y=0 into it and factorise it to get the values of x, and to find the y coordinate you sub x=0 into the equation?
    Yes, although when you substitute y=0 you won't always be able to factorise it, but you just solve the quadratic.
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: October 16, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

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