exam 1

[Jdog]

Matrices will definately be on exam one this time

what makes you so sure?

Its the first time theyve been allowed into exam 1, and theyve given us practice questions and its almost guaranteed they will test it .

[sajib_mostofa]

Then again, I don't think it will stop people from still using tree diagrams.

[pirocan1]

It would be pretty unfair of them to put it on the exam.

Is linear approximation going to be on exam 1? Cause our teacher kind of skipped it even though it was an exam question last year.

[Jdog]

yeah they can test linear approx.

[8039]

It would be pretty unfair of them to put it on the exam.

Is linear approximation going to be on exam 1? Cause our teacher kind of skipped it even though it was an exam question last year.

Many students stuff up linear approximation so I bet they'll include it somewhere

[Martoman]

Linear approx was tested for understanding last year. You most definately need to be able to work with it manually. I would expect more Yr11 probability questions which tend to be less formula based as well as some graphs... there were none last year.

Exam 1 tests your ability to make stupid mistakes. Double thinking what you are doing does help with catching out your mistakes.

[stonecold]

Linear approx was tested for understanding last year. You most definately need to be able to work with it manually. I would expect more Yr11 probability questions which tend to be less formula based as well as some graphs... there were none last year.

Exam 1 tests your ability to make stupid mistakes. Double thinking what you are doing does help with catching out your mistakes.

I haven't looked at that linear approximation question for VCAA 2009, but for the worded part, was it along the lines of if the gradient of the curve is decreasing after the point of tangency, the approximation will be greater than the actual value, whilst if the gradient of the curve is increasing after the point of tangency, the approximated value will be less than the actual value?

Edit:  These conclusions actually depend on whether the function is increasing or decreasing too yeah.  I just realised that what I said is for an increasing function I think...

[m@tty]

Geometrically, linear approximation uses the gradient from a single point to create a tangent, and then approximates values from this tangent.

If the gradient of the function is increasing, then the tangent lies below the function and hence the approximation is lower than the actual value. This is certain as long as the gradient of the function continues increasing.

Similarly, if the gradient of the function is decreasing, then the tangent lies above the function and hence the approximation is greater than the actual value. This is certain as long as the gradient of the function continues decreasing.

It was best to draw a diagram for this: the function (which was ), and the tangent at x=2, then you can clearly show that the approximated value is greater than the actual value at x=2.05(from memory).

[sajib_mostofa]

Geometrically, linear approximation uses the gradient from a single point to create a tangent, and then approximates values from this tangent.

If the gradient of the function is increasing, then the tangent lies below the function and hence the approximation is lower than the actual value. This is certain as long as the gradient of the function continues increasing.

Similarly, if the gradient of the function is decreasing, then the tangent lies above the function and hence the approximation is greater than the actual value. This is certain as long as the gradient of the function continues decreasing.

It was best to draw a diagram for this: the function (which was ), and the tangent at x=2, then you can clearly show that the approximated value is greater than the actual value at x=2.05(from memory).

Perfect explanation