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)

Related Further Mathematics GCSE answers

All answers ▸

Finding the derivative of a polynomial.


How do I find the limit as x-->infinity of (4x^2+5)/(x^2-6)?


Plot the graph of 1/x for x greater than 0.


3x^3 -2x^2-147x+98=(ax-c)(bx+d)(bx-d). Find a, b, c, d if a, b, c, d are positive integers


We're here to help

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