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# Help with Coordinate Geometry question in AS Maths? watch

1. I have no idea where to even start with this question?
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2. (Original post by jessnoch)
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.
3. (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
4. (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?

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Updated: October 17, 2015
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