Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 56

(a) Take the derivative. Use the product rule.
y'=(x-2)(2x+3)+(x^2+3x)(1)= (x-2)(2x+3)+(x^2+3x)
Given the point (1,-4) , substitute x=1 .
y'(1)= (1-2)(2(1)+3)+(1^2+3(1)) = -1(5)+(4)=-1
The slope at is -1.
Write the slope intercept form, substitute the point and the slope to find b.
y=mx+b
-4=(-1)(1)+b
b=-3
The equation of the tangent line is: y=-x-3
(b) Graph: This is only the original function... The graph will not allow me to plot y=-x-3 to add on this graph, but this function should show the tangent line on the graph. I have attached another image to the answer.

(c) You can do this on your own. It's the dy/dx function.