Question

How do you find the equation to the tangent of the curve `y= -x^3+6x^2-5x` through point A(1,0)?

Answer 1

First evaluate the derivative:
`y'(x)=-3x^2+12x-5`
Then evaluate it at `x=1`:
`y'(1)=-3+12-5=4` which represent the slope `m` of the tangent line.
Finally use the relationship:
`y-y_0=m(x-x_0)` to find the equation of the tangent line, using the coordinates of your point:
`y-0=4(x-1)`
`y=4x-4`