Year 13 Maths Help Thread

It's a basic graph translation, so the roots are translated by the same amount in the same direction. X+1 means they are translated by vector [-1,0] and so are the roots

I have summer work, can someone help Find the equation of the normal to y=2 (root x) + 8/x at the point (4, 6)I have the answers, and it says x=4 but I keep getting y=6 ????? Thanks

Are you finding the tangent instead of the normal?

No because I get the gradient as 0 and to find the perpendicular 0*-1 = 0

I got 0 by simplifying the equation to: y=2x^1/2 + 8x^-1

Then Differentiation: dy/dx = x^-1/2 - 8x^-2

and then sub 4: dy/dx = 4^-1/2 -8(4)^-2

so: dy/dx = 0 perpendicular: -1*0 =0
y-y^1 = m(x-x^1)

so: y-6=0(x-4)

which equals y=6

This thread is for people who are studying A2 Level Maths and Further Maths and is the successor of the Year 12 Maths Help Thread which was created last year. Although I hope to be relaxing much of the time, I understand that other people want to keep their maths skills refreshed over the summer.

The rules are the same as the Year 12 Maths Help Thread:

Rules

1. Keep the discussion related to maths although I do not mind the occasional physics query.
2. Please post your full working as this will enable a helper to detect any errors in your method.
3. If you want to help somebody, don't post full solutions. Your help is meant to act as a springboard, not a featherbed as giving full solutions is unlikely to actually make somebody learn.

Helpers:
If you want to be a helper, please state your intention to become one and place a brief reason why.

List:

Ano123
B_9710
SeanFM
metrize
RDKGames
Zacken

Relevant Links:

I have summer work, can someone help Find the equation of the normal to y=2 (root x) + 8/x at the point (4, 6)I have the answers, and it says x=4 but I keep getting y=6 ????? Thanks

No because I get the gradient as 0 and to find the perpendicular 0*-1 = 0

I got 0 by simplifying the equation to: y=2x^1/2 + 8x^-1

Then Differentiation: dy/dx = x^-1/2 - 8x^-2

and then sub 4: dy/dx = 4^-1/2 -8(4)^-2

so: dy/dx = 0 perpendicular: -1*0 =0
y-y^1 = m(x-x^1)

so: y-6=0(x-4)

which equals y=6

When a line has a gradient of 0, it is parallel to the x axis. The normal to this, which is perpendicular would then be parallel to the y axis, this gradient tends to infinity. So you're right up until you mutliplied the gradient by -1, think about it, the tangent is not going to have the same gradient as the normal is it?

oohhhh yeah I get the concept now.But how would I show that in my working out to finally get the answer x=4

oohhhh yeah I get the concept now.But how would I show that in my working out to finally get the answer x=4

You could Just write gradient = $\infty$ so vertical line. equation is x=4, the person marking would assume, you knew what you were doing. For your own benefit drawing a sketch would help, it doesn't matter if it's the actual curve, just the important bits need to help you visualise it.

Well I'm unsure how you would show it mathematically due to division by 0 for the normal but I would say that the tangent is completely horizontal thus y=6 therefore the normal is completely vertical and must go through (4,6) therefore X=4

Prove that the equation $x^3-47x^2+291=0$ has no integer solutions
Prove further that the equation has only irrational roots..
[Note. No credit is given for using the rational roots theorem.]