##### Differentiation, Derivatives, gradient, tangent – IB Mathematics SL

Find the equation of the tangent to the curve with equation

\( f(x)=2x^3+5x^2-20x+1\) at the point \((0,1)\)

*Solution*

The gradient of the curve at \((0,1)\) is

\(f’(0)=6(0)^2+10(0)-20=-20 \)The equation of the tangent on this point is given by the equation

\(y-y_{0}=m_{t}(x-x_{0})\)where \( m_{t}=-20 \) is the gradient of the tangent

and the touch point is \((x_{0},y_{0})=(0,1)\).

So, the equation of the tangent at the given point is

\(y-1=-20(x-0) \Rightarrow y=-20x+1\)