Calculus problem: FInd the slope of the tangent line to the parabola

Find the slope of the tangent line to the parabola y=x^2+2x at the point (-3,3) using definition 1 and 2 of limits.
Def 1 f(x)-f(a)/(x-a)
Def 2 f(a+h)-f(a)/h

Using Def 1:
Slope s = {f(x) - f(a)}/(x-a) where a is very near to x at (-3, 3)
We can denote this as a function of a: s(a) = {f(x) - f(a)}/(x-a) where a is very near to x at (-3, 3).
We have:
s = {x^2 + 2x - a^2 - 2a}/(x - a)
= {(x + a)(x - a) + 2(x - a)}/(x - a)
= x + a + 2, the idea is always to cancel out anything that is very near to zero!
In the limit when a approaches x=-3
s = -6 + 2 = -4

Using Def 2:
Slope s = {f(a + h) - f(a)}/h where h approaches 0 when a = -3
We can denote this as a function of h: s(h) = {f(a + h) - f(a)}/h where h is very near to 0.
We have:
s = {(a + h)^2 + 2(a + h) - a^2 - 2a}/h
= (a^2 + 2ah + h^2 + 2a + 2h - a^2 - 2a)/h
= (2ah + h^2 + 2h)/h
= 2a + h + 2, the idea is always to cancel out anything that is very near to zero!
In the limit when h approaches 0 at a = -3:
s = -6 + 2 = -4

-4