A) Tangent slope is f'(3)=2ax+b=6a+b when x=3.
B) and C) y=4-2x has a slope of -2=6a+b. The equation of the tangent is y=(6a+b)x+c where c has to be found by plugging in (3,f(3))=(3,(9a+3b)): 9a+3b=3(6a+b)+c; 9a+3b=18a+3b+c, so c=-9a and the equation of the tangent line at x=3 is y=(6a+b)x-9a. Comparing this with y=-2x+4, 9a=-4, a=-4/9 and 6a+b=-8/3+b=-2, b=8/3-2=2/3. y=f(x)=-4x^2/9+2x/3 or y=2x/3-(2x/3)^2. y'=2/3-8x/9. When x=3, y'=2/3-8/3=-2. So f(x) is confirmed with the values of a and b we found. The tangent line is also confirmed by inserting values for a and b in y=(6a+b)x-9a=(-8/3+2/3)x+4=-2x+4.
D) f(4)=-4*16/9+8/3=-64/9+8/3=-40/9=-4.44 approx. y=4-2*4=-4. So the equation of the tangent contains the point (4,-4) while the curve of f(x) contains the point (4,-4.44) which is close to (4,-4) with an error of 4/9. This means that at the point where x=4 (draw a vertical line to pass through the tangent line and the curve), the parabolic curve is slightly below the line (steeper), but the curve is quite close to being a straight line between x=3 and 4.