Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    Name:  parabola graph.png
Views: 85
Size:  53.7 KBFor this graph I have to find the equation in the form y = ax2 + bx + c but i don't know how to answer it .
    Is it completing the square?
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by ChocoholicCox)
    Name:  parabola graph.png
Views: 85
Size:  53.7 KBFor this graph I have to find the equation in the form y = ax2 + bx + c but i don't know how to answer it .
    Is it completing the square?
    You know two points on the curve (0.10) and (3,1) and you know that there is a minimum turning point when x=3.
    This information should enable you to write down three simultaneous equations in a,b and c
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    You know two points on the curve (0.10) and (3,1) and you know that there is a minimum turning point when x=3.
    This information should enable you to write down three simultaneous equations in a,b and c
    I swear all you have to do is (x+3)2+ 1 = 0
    then just multiply the brackets giving x2 + 6x + 9 then just plus the 1 giving x2 + 6x +10 because the x2​ shows its a parabola?
    Offline

    0
    ReputationRep:
    (Original post by ChocoholicCox)
    I swear all you have to do is (x+3)2+ 1 = 0
    then just multiply the brackets giving x2 + 6x + 9 then just plus the 1 giving x2 + 6x +10 because the x2​ shows its a parabola?
    Wouldn't it be (x-3)^{2}+1=0 ? As 3 is the minimum (what makes the bracket equal 0)

    You could also note right off the bat that c=10 as the graph crosses the y-axis at (0,10)
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by Robbie242)
    Wouldn't it be (x-3)^{2}+1=0 ? As 3 is the minimum (what makes the bracket equal 0)

    You could also note right off the bat that c=10 as the graph crosses the y-axis at (0,10)
    Yes, you can simply think of it as the standard parabola y=x^2, translated 3 units to the right and 1 unit upwards.
    Offline

    0
    ReputationRep:
    (Original post by brianeverit)
    Yes, you can simply think of it as the standard parabola y=x^2, translated 3 units to the right and 1 unit upwards.
    Yeah I know don't worry, I was just questioning the OPs deduction
    Offline

    1
    ReputationRep:
    (Original post by ChocoholicCox)
    I swear all you have to do is (x+3)2+ 1 = 0
    then just multiply the brackets giving x2 + 6x + 9 then just plus the 1 giving x2 + 6x +10 because the x2​ shows its a parabola?
    Almost...
    \boxed{y=a(x-h)^2+k}
    You have a vertex @ (3,1) and when x=0 y=10
    The boxed equation is the quadratic equation in vertex form, where (h, k) = (3, 1)
    So: Sub your values in: 10=a(0-3)^2+1
    10=9a+1
    9=9a
    a=1
    Substitute a back into your quadratic equation in vertex form:
    y=1(x-3)^2+1
    Expand: y=x^2-6x+10

    Can you see how you need to know what a is before you go ahead and use that vertex equation? It just so happens that the y-intercept is 9, but if it was 4.5 or 20, then your value of a would change and you'd have a slightly different equation...
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by halpme)
    Almost...
    \boxed{y=a(x-h)^2+k}
    You have a vertex @ (3,1) and when x=0 y=10
    The boxed equation is the quadratic equation in vertex form, where (h, k) = (3, 1)
    So: Sub your values in: 10=a(0-3)^2+1
    10=9a+1
    9=9a
    a=1
    Substitute a back into your quadratic equation in vertex form:
    y=1(x-3)^2+1
    Expand: y=x^2-6x+10

    Can you see how you need to know what a is before you go ahead and use that vertex equation? It just so happens that the y-intercept is 9, but if it was 4.5 or 20, then your value of a would change and you'd have a slightly different equation...
    oh right i see what u mean
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 10, 2013
  • create my feed
  • edit my feed

University open days

  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 23 Nov '18
  • Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 23 Nov '18
  • Edge Hill University
    All Faculties Undergraduate
    Sat, 24 Nov '18
Poll
Black Friday: Yay or Nay?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.