Differentiation - Chain Rule (1 Viewer)

[skillstriker]

Differentiate y = a(x+b)² - 8, then find 'a' and 'b' if the parabola:

1. passes through the origin with gradient 16
2. has tangent y=2x at the point P (4,8)

Thanks!

 

[Carrotsticks]

For Q1, we sub in the conditions P(0) = 0 and P'(0) = 16:



For Q2 do the exact same thing, except we sub in the conditions P(4) = 8 and P'(4) = 2

 

[skillstriker]

Thanks. I'm stuck on another question.

a) Find the equation of the tangent to y = 1/(x-4) at the point L where x=b
- I got the answer for this as x+y(b-4)² = 2b-4 which is correct (confirmed by textbook)

However, I'm stuck on the second part of this question, which is:

b) Hence find the equation of the tangent to the curve passing though

1. the origin
2. W (6,0)

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Equation of tangent is }x@plus;y(b-4)^2=2b-4\\ \textup{Passes through the origin, therefore 0,0 satisfies this:}\\ 0@plus;0(b-4)^2=2b-4\\ b=2 \textup{ Sub back into tangent:}\\ x@plus;y(2-4)^2=2(2)-4\\ x@plus;4y=0 \\ \textup{Use a similar process for the other part}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Equation of tangent is }x+y(b-4)^2=2b-4\\ \textup{Passes through the origin, therefore 0,0 satisfies this:}\\ 0+0(b-4)^2=2b-4\\ b=2 \textup{ Sub back into tangent:}\\ x+y(2-4)^2=2(2)-4\\ x+4y=0 \\ \textup{Use a similar process for the other part}" title="\textup{Equation of tangent is }x+y(b-4)^2=2b-4\\ \textup{Passes through the origin, therefore 0,0 satisfies this:}\\ 0+0(b-4)^2=2b-4\\ b=2 \textup{ Sub back into tangent:}\\ x+y(2-4)^2=2(2)-4\\ x+4y=0 \\ \textup{Use a similar process for the other part}" /></a>