Given ƒ(x) = 2x^2 - 3x - 1 find the slope and equation of the tangent line to ƒ at the point (1, -2)

Answer 1

dy/dx = 4x - 3

at (1,-2), dy/dx = slope = 4(1)-3 = 1

so using y = mx + b
-2 = (1)(1) + b
b = -3

y = x - 3 is the tangent equation.