Find the equation of the tangent to the curve y = 2x^2 + x - 1 at the point where x = 1.

To get the required tangent equation we need its gradient and the coordinates of a point it passes through. We can then substitute it into the formula y - y1 = m(x - x1). 1. Finding the gradient: To find the gradient of the tangent, first find the gradient of the curve where x=1: y = 2x2 + x - 1 --> dy/dx = 4x + 1 When x = 1, dy/dx= 4(1)+1 = 4+1 = 5 So the gradient of the required tangent is 5. 2. Finding a point the tangent passes through: We know that the tangent touches the curve when its x coordinate is 1, and we can also calculate the y coordinate of this point: y = 2(1)2 + (1) - 1 = (as y = 2x2 + x - 1). 3. Substiuting this into the line equation formula and rearranging: We can use the line equation formula: y - y1 = m(x - x1) and subsitute in the information we now have (m=5, x1=1 and y1=2). y - 2 = 5(x - 1) y - 2 = 5x - 5 --> y = 5x - 3 So y = 5x - 3 is the equation of the tangent to the curve y = 2x2 + x - 1 at the point where x = 1.

BT
Answered by Beth T. Maths tutor

11403 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.


given that at a time t, a particle is accelerating in the positive x-direction at 1/t ms^-2, calculate the velocity and the displacement of the particle at time t = 2s


find the diffrential of 3sin2x+4cos2x


Given that y > 0, find ∫((3y - 4)/y(3y + 2)) dy (taken from the Edexcel C4 2016 paper)


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning