Finding the equations of tangent line and normal line

Question

Find the equations of the tangent line and normal line to the curve y=2x³- e^x, at x=0

Answer 1

At x=0, y=-1

dy/dx=6x²-e^x which is -1 when x=0

Use the point slope form:

The equation of the tangent line is -1=(y+1)/x

The slope of the normal (perpendicular to the tangent) is 1

and the equation of the normal is 1=(y+1)/x.