First, find the derivative of #f(x)#

#f'(x) = -3+10(x-4)-6(x-4)^2+24(x-4)^3#

Next, find the second derivative of #f(x)#

#f''(x)=10-12(x-4)+74(x-4)^2#

The formula for finding the Taylor series expansion of a function is

#f(a) + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f^(3)(a))/(3!)(x-a)^3+...#

Plugging in up to the second derivative term is all that is necessary because we only want the 2nd degree Taylor:

#f(4)+(f'(4))/(1!)(x-4)^1 + (f''(4))/(2!)(x-4)^2#

Lets get each of the terms #f(4)#, #f'(4)#, and #f''(4)#

#f(4) = 7-3(0)+5(0)^2-2(0)^3+6(0)^4=7#

#f'(4) = -3+10(0)-6(0)^2+24(0)^3 = -3#

#f''(4) = 10-12(0)+74(0)^2 = 10#

Plugging these terms in to the above Taylor series gives

#7-3(x-4)^1 + 5(x-4)^2#

If you had been asked to find the approximation of #f(x)# at #x=4.3#, you could now just plug in #x=4.3# into this, giving the approximation:

#f(4.3) ~~ 7-3(4.3-4)+5(4.3-4)^2= 6.55#

But Nay! You were asked to approximate #f'(x)# at #x=4.3#, not #f(4.3)#. At least that's the question above. So, starting with

#f'(x) = g(x) = -3+10(x-4)-6(x-4)^2+24(x-4)^3#

#g'(x) = 10-12(x-4)+72(x-4)^2#

#g''(x) = -12+144(x-4)#

At #x=4#, these give

#g(4) = -3#

#g'(4) = 10#

#g''(4) = -12#

Plugging *these* values in, gives

#g(4)+(g'(4))/(1!)(x-4)^1 + (g''(4))/(2!)(x-4)^2#

#=-3+(10)/(1!)(x-4)^1 + (-12)/(2!)(x-4)^2#

#=-3+10(x-4) -6(x-4)^2#

So, the final answer is

#g(4.3)=f'(4.3)~~-3+10(4.3-4) -6(4.3-4)^2=-0.54#