y = 2x^2 - 13x + 5
dy/dx = 4x - 13
At x = 3, dy/dx = 4(3) - 13 = 12 - 13 = -1
and y = 2(3)^2 - 13(3) + 5 = 2(9) - 39 + 5 = 18 - 34 = -16
The tangent line at x = 3, passes through the point (3, -16) and has a slope of -1
y - (-16) = -1(x - 3)
y + 16 = -x + 3
y = -x - 13 is the required equation.
