A curve has equation y = x^2 - 7x. P is a point on the curve, and the tangent to the curve at P has gradient 1. Work out the coordinates of P.

The gradient of the tangent at P (=1) will be equal to the gradient of the curve at point P (=1). The gradient at a point on a curve is given by dy/dx (the differential). First, differentiate the equation of the curve, and equate the differential to 1. Solve for x by rearranging the equation. Finally, plug this value of x into the equation for y, and solve for y. dy/dy = 2x - 7 = 1, 2x = 8, x = 4. Therefore, y = (4)^2 - 7*(4) = -12Thus, the coordinates of P are (4,-12)

ST
Answered by Serene T. Further Mathematics tutor

2278 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Point A lies on the curve: y=x^2+5*x+8. The x-coordinate of A is -4. What is the equation of the normal to the curve at A?


l1 and l2 are tangents of a circle. l1 intersects the circle at (3-√3,5) with a gradient of √3, and l2 intersects the circle at (3+√2,4+√2) with a gradient of -1. Find the centre of the circle, and hence find the radius of the circle.


A curve has equation: y = x^3 - 3x^2 + 5. Show that the curve has a minimum point when x = 2.


Find the solution of 3^{4x} = 9^{(x-1)/2}.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences