Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Help with Coordinate Geometry question in AS Maths? watch

    • Thread Starter
    Offline

    1
    ReputationRep:
    I have no idea where to even start with this question?
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by jessnoch)
    I have no idea where to even start with this question?
    Do you know how to sketch the graphs? work out where the first curve crosses the axes, and label those points on your diagram. With the second sketch, it's just a straight line. Label where it intersects the axes in terms of k.

    For part b, you know that a tangent to a curve has the same gradient as the curve at that point, so you need to differentiate C and let this equal the gradient of your straight line, to find a point that both lines pass through.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by halfhearted)
    Do you know how to sketch the graphs? work out where the first curve crosses the axes, and label those points on your diagram. With the second sketch, it's just a straight line. Label where it intersects the axes in terms of k.

    For part b, you know that a tangent to a curve has the same gradient as the curve at that point, so you need to differentiate C and let this equal the gradient of your straight line, to find a point that both lines pass through.
    I'm unsure of how to sketch the graphs as I don't understand how you get points from the equations. I looked online for tutorials but I can't find any
    Offline

    3
    ReputationRep:
    (Original post by jessnoch)
    I'm unsure of how to sketch the graphs as I don't understand how you get points from the equations. I looked online for tutorials but I can't find any
    look for a tutorial on transformations of graphs. In this case, you've got an x^2 graph that's squished down and moved upwards. This gives you the basic shape. If you do this, you'll see that it doesn't actually cross the x-axis, as it has no real roots (1/3(x^2)+8=0 has no solutions)To find where it crosses the y-axis, find what y equals when x=0, ie just substitute x=0 into the equation of the line.For line L, you should recognise the form y=mx+c, where m is your gradient and c is your y-intercept. Here, m=3 and c=k. Now can you work out where this crosses the x-axis?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 17, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.