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