Find the tangent for the line y=x^3+3x^2+4x+2 at x=2

Firstly, differentiate the equation y=x3+3x2+4x+2 to find the gradient function. The gradient function is dy/dx and is found to be 3x2+6x+4 = f'(x) after differentiation. The simple rule is to multiply the term by the power of x, and then subtract one from the power to differentiate each term. Also you should remember to remove the constant when differentiating (in this case "2") because the power of x is 0 and multiplying a term by 0 gives 0. Then, to find the gradient at the point x=2, simply put the number 2 into the gradient function to find the gradient. So, the gradient m=f'(2) = 28. To find the tangent, first find the y coordinate where x=2 by substituting this into the original equation (the first equation). This is found to be 30. So the coordinates are (2,30) of where the tangent will cross. Now, apply the equation y-y1=m(x-x1), where (x1,y1) are the coordinates of the tangent. The equation looks like this: y-30=2(x-28) This turns out to be, after rearranging, y=2x-26.

WC
Answered by Wafi C. Maths tutor

5008 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is the moment about the pivot C


How would you derive y = function of x; for example: y = 3x^3 + x^2 + x


The curve y = 2x^3 -ax^2 +8x+2 passes through the point B where x = 4. Given that B is a stationary point of the curve, find the value of the constant a.


The curve with the equation: y=x^2 - 32sqrt(x) + 20 has a stationary point P. Find the coordinates of P.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning