VCE Methods Question Thread!

[Lear]

https://youtu.be/OkFdDqW9xxM

Have a watch of this, helped me a lot

[sailinginwater]

https://youtu.be/OkFdDqW9xxM

Have a watch of this, helped me a lot

Thanks
2014 methods exam 1
Question 10a
I substituted the point (2,4) into
y = ax^2 + bx and thus got 4a + 2b = 4
I'm not sure how to go any further with this question. Can someone please help?

[Lear]

Also note that since the line QU is a tangent to y, the gradient of QU is the same as the gradient of y at x=2.

[sailinginwater]

Also note that since the line QU is a tangent to y, the gradient of QU is the same as the gradient of y at x=2.

Does this rule apply to all tangents at a particular point?

[Lear]

Does this rule apply to all tangents at a particular point?

As per the textbook definition
'The tangent line to the graph of the function f at the point (a, f (a)) is defined to be the
line through (a, f (a)) with gradient f'(a).'

[sailinginwater]

As per the textbook definition
'The tangent line to the graph of the function f at the point (a, f (a)) is defined to be the
line through (a, f (a)) with gradient f'(a).'

Thanks
I found the gradient of the parabola to be 2ax + b
And at x = 2, 4a + b
But how do we now the gradient if the the straight line when we only have (2,4) as complete cordinates?

[Lear]

Thanks
I found the gradient of the parabola to be 2ax + b
And at x = 2, 4a + b
But how do we now the gradient if the the straight line when we only have (2,4) as complete cordinates?

You also have U (the x intercept of the tangent)

[sailinginwater]

You also have U (the x intercept of the tangent)

But it says when u = 6, so doesn't that indicate that u can have more than one value?

[Lear]

Yes but when U=6 it is tangent. So you know that when U=6 the gradient at that point is the same as the gradient between given coordinates and (6,0)

[sailinginwater]

Yes but when U=6 it is tangent. So you know that when U=6 the gradient at that point is the same as the gradient between given coordinates and (6,0)

Thanks
Methods exam 1 2014 question 10bii

Find the minimum total shaded area and the value of u for which the area is a minimum

I found the area of the shaded region to be (4u^2)/(u-2) – 8 and differentiated it to be

(4u^2-16u)/((u-2))^2

Then let (4u^2-16u)/((u-2))^2 = 0, and I got u = 0 and u = 4 but how do we know which value of u to use?

[Lear]

Just sub them in and observe what the area is.

[sailinginwater]

Just sub them in and observe what the area is.

What's the difference between finding the minimum and maximum area?

[Lear]

Uh...the minimum area is the lowest value of the area while maximum is the highest value.

[sailinginwater]

Uh...the minimum area is the lowest value of the area while maximum is the highest value.

The above question I posted said the endpoints were u = 5/2 and u = 6
Do you know where they got the endpoints from?

[S_R_K]

The above question I posted said the endpoints were u = 5/2 and u = 6
Do you know where they got the endpoints from?

It's stated at the beginning of the question that 5/2 ≤ u ≤ 6.