Find the gradient of the tangent to the line y=(x-2)^2 at the point that it intercepts the y-axis

First find the coordinates of the point in question:We know x=0By plugging this into the equation of the line we get y=(0-2)2 = (-2)2 = 4Therefore the point is (0,4)
To find the gradient of a line, we differentiate the equation of the line:By substitution -> y=u2 , u=x-2dy/dx=dy/du.du/dxdy/du = 2u , du/dx=1Therefore dy/dx =2u=2x-4Subbing in known coordinate into this equation we get:dy/dx(x=0,y=4) = -4Answer = -4

AJ
Answered by Alec J. Maths tutor

3440 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A level Maths question - The graph of y=2sin(2x)+1 is rotated 360 degrees about the x-axis to form a solid. Find the volume enclosed by the curve, the co-ordinate axes and the line x=pi/2


Prove that the indefinite integral of I = int(exp(x).cos(x))dx is (1/2)exp(x).sin(x) + (1/2)exp(x).cos(x) + C


A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.


Simplify the following expression to a fraction in its simplest form: [(4x^2 + 6x)/(2x^2 - x -6)] - [(12)/(x^2 - x - 2)]


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