A curve has equation y=(2-x)(1+x)+3, A line passes through the point (2,3) and the curve at a point with x coordinate 2+h. Find the gradient of the line. Then use that answer to find the gradient of the curve at (2,3), stating the value of the gradient

The gradient of the line is given by change in Y divided by change in X. To find the Y value of the point we must put 2+h into the y=(2-x)(1+x)+3. Doing this we get y=(2-(2+h))(1+(2+h))+3=(-h)(3+h)+3= -3h -h2 +3.

So the change in Y is from 3 to (-3h -h2 +3), the change between these two values is 3h+h2) (found by subtracting the second from the first, remember the order must be the same for Y and X)

The change in x is simply the second subtracted from the first (like with the Y values), so the change in x = 2 -(2+h) = -h

So the gradient is (3h+h2)/-h = -h-3

We can use this to find the gradient at (2,3), if we imagine h getting smaller and smaller (tending to 0) then the line of (2,3) to the point with x coordinate 2+h becomes a line from (2,3) to a point incredibly close to (2,3), so close it tends to it and we can say it is (2,3), not just its close anymore.

So as h tends to 0, the gradient of the line in the first part becomes the gradient of the curve at (2,3), so the gradient is -h-3, as h tends to 0 the gradient becomes -3

JC
Answered by James C. Further Mathematics tutor

2578 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Write 1 + √3i in modulus-argument form


What does it mean if two matrices are said to be commutative?


Find the general solution to the differential equation y'' + 4y' + 3y = 6e^(2x) [where y' is dy/dx and y'' is d^2 y/ dx^2]


A golf ball is hit from horizontal ground with speed 10 m/s at an angle of p degrees above the horizontal. The greatest height the golf ball reached above ground level is 1.22m. Model the golf ball as a particle and ignore air resistance. Find p.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences