You can use the definition:
[tex]\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
Then if
[tex]f(x) = 2x^2+3x-2[/tex]
we have
[tex]f(x+h) = 2(x+h)^2+3(x+h) - 2 = 2x^2 + 4xh+2h^2+3x+3h-2[/tex]
Then the derivative is
[tex]\displaystyle f'(x) = \lim_{h\to0}\frac{(2x^2+4xh+2h^2+3x+3h-2)-(2x^2+3x-2)}h \\\\ f'(x) = \lim_{h\to0}\frac{4xh+2h^2+3h}h \\\\ f'(x) = \lim_{h\to0}(4x+2h+3) = \boxed{4x+3}[/tex]
I'm guessing the second part of the question asks you to find the tangent line to f(x) at the point a = 0. The slope of the tangent line to this point is
[tex]f'(0) = 4(0) + 3 = 3[/tex]
and when a = 0, we have f(a) = f (0) = -2, so the graph of f(x) passes through the point (0, -2).
Use the point-slope formula to get the equation of the tangent line:
y - (-2) = 3 (x - 0)
y + 2 = 3x
y = 3x - 2
