The curve C has a equation y=(2x-3)^5; point P (0.5,-32)lies on that curve. Work out the equation to the tangent to C at point P in the form of y=mx+c

Firstly you should work out the first derivative of the equation y= After differentiating the equation, sub the x value of the point P into the first derivative. This should give you the gradient of the equation. After getting the gradient of the tangent, you could use the y and x value of point P and sub it into the equation of y=mx+c to work out the y intercept (c). This would give you the answer.

MR

Related Maths A Level answers

All answers ▸

A curve has equation x^2 + 2xy – 3y^2 + 16 = 0. Find the coordinates of the points on the curve where dy/dx =0


Solve the following pair of simultaneous equations: 2x - y = 7 and 4x + y = 23


Given that y=x^3 +2x^2, find dy/dx . Hence find the x-coordinates of the two points on the curve where the gradient is 4.


Differentiate 3x^2+1/x and find the x coordinate of the stationary point of the curve of y=3x^2+1/x