Question

The equation of the line tangent to the curve y = (kx+8)/ (k+x) at x=-2 is y=x+4. What is the value of k?

Answer 1

differentiate using the quotient rule to find the equation of the gradient

dy/dx = ((k + x)1 - ( k+x)k)/(k +x)^2

its known from the information the gradient is m = 1 when x = - 2

then 1 = (( k - 2) - (k - 2)k)/(k - 2)^2

simplifying

1 = -(k -2)(k - 1)/(k-2)^2

1 = (k - 1)/(k - 2)

or k - 2 =- k + 1

2k = 3

k = 3/2

the value of k is 3/2

Answer 2

Well this was a multiple choice question and i wanted to know the steps but 3/2 isnt an answer choice....the choices are -3, -1, 1, 3, and 4 and he answer is 3

Answer 3

but im not sure whats wrong with ur work

Answer 4

its in the derivative it should be

dy/dx = ((k + x)k) - (kx + 8)1)/(k + x)^2

dy/dx = (k^2 - 8)(k + x)^2

sorry about that

Answer 5

oh no its ok.....can u kind of explain though why u found the derivative and what not

Answer 6

oops dy/dx = (k^2 - 8)(k +x)^2

Answer 7

wait shud it be a divison sign btwn the 2 parenthesis....

Answer 8

the deriviative gives the equation of the gradient of the tangent. By substituting the x value,
x = -2 you will get the gradient at a specific point on the curve

derivative is dy/dx = (k^2 - 8)/(k + x)^2

Answer 9

then tangent equation has a gradient of 1 so evaluate the derivative dy/dx = 1 when x = -2 to find k

so 1 = (k^2 - 8)/(k - 2)^2


or (k - 2)^2 = k^2 - 8
expanding gives

then k^2 - 4k + 4 = k^2 - 8

-4k = -12

k = 3

Answer 10

THANKS SO MUCH....I APPRECIATE IT =)

Answer 11

especially u explaining it and showing it step by step....thanks 4 teh patience