Show an equation of the tangent to the curve π¦ = π₯^5 β 3π₯^2 +π₯ + 5 at the point (β1, 0) is given by π¦ = 12(π₯ +1) by differentiation
Show an equation of the tangent to the curve π¦ = π₯^5 β 3π₯^2 +π₯ + 5 at the point (β1, 0) is given by π¦ = 12(π₯ +1) by differentiation
y' = 5x^4 - 6x + 1
The slope of the tangent line at (-1,0) is
y' (-1) = 5(-1)^4 - 6(-1) + 1 = 12
The equation of the tangent line to (-1,0) is
y = 12 ( x - 1) + 0
y = 12(x - 1)