Find an equation of the tangent line to the curve at the given point.
y =x^4+4x^2−x,(1,4)
y=?
well, y' = 4x^3 + 8x - 1
So, at (1,4), the slope of the tangent is 11.
That means the line there is
y-4 = 11(x-1)
y =x^4+4x^2−x,(1,4)
y=?
So, at (1,4), the slope of the tangent is 11.
That means the line there is
y-4 = 11(x-1)