Question

Find the equation of the tangent lines at a point (1, 2) of the curve
y=x3+1​

Answer 1

Find

dy/dx of y equation

y=3x2−x3

dy/dx=6x−3x2

Then from the point (1,2) we can obtain the x value which is 1

Insert

x=1

into

dy/dx

dy/dx=6(1)−3(1)2

dy/dx=3

To form the equation we have to base it from the original equation

y=mx+c

we know that m=3 from

dy/dx=3

y=3x+c

use the coordinates given fro x and y values to find c

(1,2) x=1 , y=2

(2)=3(1)+c

c=−1

Then we put c back into the general form to get the final tangent equation

y=3x−1