Find the equation of the tangent to the curve y = 3x^2 + 4 at x = 2 in the form y = mx + c

There are two main steps. First find the gradient of the curve at x = 2 (m). This is done by differentiating the curve equation y = 3x^2 + 4 to get dy/dx = 6x. By plugging in x = 2, we get the gradient of the tangent, m, as 62 = 12. Then we need to find the y intercept of the tangent, c. We make c the subject, so c = y - mx. We worked out what m is (12) so we just need a set of coordinates x,y which lie on the tangent. The easiest point is where the tangent meets the curve. We know x = 2 so plugging that into the curve equation gives y = 3(2^2) + 4 = 16. Now we have values of x,y,m we can find c. c = 16 - 12*2 = -8. Therefore the final answer is y = 12x - 8

Related Maths A Level answers

All answers ▸

A straight line passes through the point (2,1) and has a gradient of 3. Find the co-ordinates of the points where this line intersects the axes


Find the turning points of the curve y = 3x^4 - 8x^3 -3


Given y=(1+x^3)^0.5, find dy/dx.


In this question, take 'log' to mean 'log base 5'. Solve the equation log(x^2-5)-log(x) = 2*log(2)