Question

Help with question 6ii pleaseee!
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w06_qp_3.pdf

Answer 1

Parallel to x-axis implies \(\dfrac{dy}{dx}=0\).

Answer 2

What's next?

Answer 3

you get an equation
solve that simultaneously with the equation of given implicit curve to get the points at which dy/dx=0

Answer 4

from the previous ques eh?

Answer 5

what previous ques ?

Answer 6

uh, idk where to start. :(

Answer 7

\[
\frac{dy}{dx}=\frac{y-x^2}{2y^2-x}=0
\]

Answer 8

Play around with it and see what you can get.

Answer 9

y=x^2? D:

Answer 10

Put \(y=x^2\) back to the equation of the curve.

Answer 11

why?

Answer 12

Because then you can solve for the x-coordinates of points such that the tangent of the curve is parallel to x-axis.

Answer 13

i got x^3 + 2x^6 - 3x^2=0

Answer 14

Are you sure?

Answer 15

no?

Answer 16

The term on the right hand side is 3xy.

Answer 17

uhhuh?

Answer 18

so x^3 + 2x^6 -3x^3?

Answer 19

Yes, solve for x then.

Answer 20

im so confused by the powers like when to plus or multiply gosh

Answer 21

\[a^{m+n} = a^m\cdot a^n\]

Answer 22

\[x^6 = x^{3+3} = x^3\cdot x^3\]

Answer 23

same integer is plus the power right?

Answer 24

\(x^n=\underbrace{x\cdot x\cdot\>\cdots\>\cdot x\cdot x}_{\text{n times}}\)

Answer 25

consider a quick example :

\(\color{blue}{a\cdot a\cdot a = a^3}\)
\(\color{purple}{a\cdot a\cdot a \cdot a \cdot a= a^5}\)

multiply them both :
\(\begin{align}\color{blue}{a\cdot a\cdot a } \cdot \color{purple}{a\cdot a\cdot a \cdot a \cdot a} &= \color{blue}{a^3}\cdot \color{purple}{a^5}\\~\\
a^8& = \color{blue}{a^3}\cdot \color{purple}{a^5}

\end{align}\)

Answer 26

thanks you all!

Answer 27

Make sure you know what is asked when having an examination. Don't answer something that is not asked.

Answer 28

OK