Turn on thread page Beta

Higher maths help please with the stupidest question watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi guys,

    I feel so stupid about this question, can some please explain to me how to find values of a and b.

    Thanks you.
    Attached Images
  1. File Type: pdf tsr1.pdf (178.0 KB, 91 views)
    Offline

    16
    ReputationRep:
    (Original post by rabia999)
    Hi guys,

    I feel so stupid about this question, can some please explain to me how to find values of a and b.

    Thanks you.
    1. You're not stupid if you're studying higher maths
    2. I wish I could be of use but I can't remember this.
    Offline

    10
    ReputationRep:
    (Original post by rabia999)
    Hi guys,

    I feel so stupid about this question, can some please explain to me how to find values of a and b.

    Thanks you.
    Right, I'm not fantastic at explaining things, or at logarithmic graphs. But I must have done something right or I wouldn't have got my Higher Maths this year. And no your not stupid because you don't understand a particular question at first, your studying one of the hardest Highers out there for a start! Anyway...

    Logarithmic functions usually pass through the point (1,0), whereas the function in your question passes through (3,0). The line x=2 is given as a little hint. It shows that the graph has been moved to the right by 2 places. Combine this with what you learnt about Graph Transformations then you can work out 'b'.
    Spoiler:
    Show


    b=2



    Finding 'a' requires you to sub in a point on the function back into the function, here the point (4,3) is given to you and you should use that. To find 'a' you need to sub in the value you found for 'b' and the x and y values from the coordinates given into the equation in the question.
    Spoiler:
    Show


    a=3

    Offline

    13
    ReputationRep:
    I couldn't find a Higher Maths forum so I'll just post it here.

    I'm finding Definite Integrals really difficult.
    Well, I think I'm doing well but I always check the answers and it's never correct.

    It's finding the area under a curve (above and below the x-axis). I was wondering if anybody could help me at all in this topic. Any help is appreciated !
    Offline

    8
    ReputationRep:
    (Original post by Ethan100)
    I couldn't find a Higher Maths forum so I'll just post it here.

    I'm finding Definite Integrals really difficult.
    Well, I think I'm doing well but I always check the answers and it's never correct.

    It's finding the area under a curve (above and below the x-axis). I was wondering if anybody could help me at all in this topic. Any help is appreciated !
    Is this for core 1, because we've just finished doing this in class so I could help
    Offline

    13
    ReputationRep:
    (Original post by ruby_zara)
    Is this for core 1, because we've just finished doing this in class so I could help
    I'm not exactly sure what core 1 includes but I'm guessing it is...
    I just want steps on how to answer those types of questions and an example would be also great !
    Offline

    2
    ReputationRep:
    You integrate as normal, raise the power by 1 and divide by the new power. once you have you your integrated function you then substitute in your limits.

    Ex.
    Name:  Screen Shot 2016-12-11 at 22.54.55.png
Views: 76
Size:  32.8 KB
    For the area under a graph, its exactly the same steps, except they might ask you to find the limits. In which case you take your y=f(x) and set it to 0 to find the roots/x-intercepts of the graph. That would be your basic questions. If the graph is a mix of under and below the x axis then you MUST do these separately, i.e. find the areas above separately from the areas below because when you find the areas below they will come out as negative. You cant have a negative area so you must make it positive and then add it to the above areas for final answer.

    Hope this helps
    Offline

    13
    ReputationRep:
    (Original post by Tayls102)
    You integrate as normal, raise the power by 1 and divide by the new power. once you have you your integrated function you then substitute in your limits.

    Ex.
    Name:  Screen Shot 2016-12-11 at 22.54.55.png
Views: 76
Size:  32.8 KB
    For the area under a graph, its exactly the same steps, except they might ask you to find the limits. In which case you take your y=f(x) and set it to 0 to find the roots/x-intercepts of the graph. That would be your basic questions. If the graph is a mix of under and below the x axis then you MUST do these separately, i.e. find the areas above separately from the areas below because when you find the areas below they will come out as negative. You cant have a negative area so you must make it positive and then add it to the above areas for final answer.

    Hope this helps
    That's perfect ! Thanks so much for the effort
 
 
 
The home of Results and Clearing

2,996

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?
Applying to university
Uni match

Find your perfect match

Our tool will help you find the perfect uni course for you

Make your revision easier

Study Help rules and posting guidelines

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.