The first step would be to differentiate the equation of the curve. The gradient of the tangent to the curve at any point is the same as that of the curve at the same point.
When you have a point on a straight line (here, the tangent) and its gradient, you can find its equation using:

y – y1 = m(x – x1)

where y1 is the y coordinate of the point, x1 is the x coordinate and m is the gradient.

(edited 10 years ago)

Reply 6

bubbles5897

OP

Original post by William Turtle

The first step would be to differentiate the equation of the curve. The gradient of the tangent to the curve at any point is the same as that of the curve at the same point.
When you have a point on a straight line (here, the tangent) and its gradient, you can find its equation using:

y – y1 = m(x – x1)

where y1 is the y coordinate of the point, x1 is the x coordinate and m is the gradient.

i suppose the gradient is 4, and when i plug the coordinates into the equation, the answer i get isn't similar to what it says at the back. The answer at the back is 4y-3x-4=0

Reply 7

davros

16

Original post by bubbles5897

i suppose the gradient is 4, and when i plug the coordinates into the equation, the answer i get isn't similar to what it says at the back. The answer at the back is 4y-3x-4=0

How did you get a gradient of 4 at the point where x = 2 ? Can you tell us what you think the derivative of

x + \dfrac{1}{x}

is?

Reply 8

William Turtle

14

Original post by bubbles5897

i suppose the gradient is 4, and when i plug the coordinates into the equation, the answer i get isn't similar to what it says at the back. The answer at the back is 4y-3x-4=0