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)
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