Question

2x^2+3x-2, a=0 How do I find the derivative?

Answer 1

You can use the definition:

$\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h$

Then if

$f(x) = 2x^2+3x-2$

we have

$f(x+h) = 2(x+h)^2+3(x+h) - 2 = 2x^2 + 4xh+2h^2+3x+3h-2$

Then the derivative is

$\displaystyle f'(x) = \lim_{h\to0}\frac{(2x^2+4xh+2h^2+3x+3h-2)-(2x^2+3x-2)}h \\\\ f'(x) = \lim_{h\to0}\frac{4xh+2h^2+3h}h \\\\ f'(x) = \lim_{h\to0}(4x+2h+3) = \boxed{4x+3}$

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

$f'(0) = 4(0) + 3 = 3$

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