The slope of the tangent line of f(x) at x=k is equal to the derivative f'(x) evalated at k:
f(x) = 
f'(x) =
f'(x) at x = -2: f'(-2) = -8(-2) + 9 = 25
The slope of the tangent line at (-2,-34) is therefore
For the equation of the tangent line, make use of the point-slope form of a line, and the given information that the line passes through (-2,-34):
y - y0 = m(x-x0)
y-(-34) = 25(x-(-2))
y+34 = 25x+50
in slope-intercept form