Calculus of a Single Variable, Chapter 2, 2.3, Section 2.3, Problem 64

Start by taking the derivative. This will require product rule and chain rule.
The product rule is: AB'+BA'
f'(x) = (x-2)(2x) + (x^2+4)(1)
f'(x)= (x-2)(2x) + x^2+4
f'(x)= 2x^2 -4x+x^2+4
The derivative is:
f'(x) = 3x^2-4x+4
Substitute the x-value of the given point.
f'(1) = 3(1)^2-4(1)+4
f'(1)=3
The slope at the point (1,-5) is 3.
The equation of the tangent line is in the form of:
y=mx+b
Substitute the slope, 3, and the given point, (1,-5), into the equation and find the y-intercept, b.
-5=(3)(1)+b
-5=3+b
b=-8
Substitute the slope and y-intercept back into the equation of the tangent line.
a)
y=3x-8
Graph the original function with the equation of the tangent line. See image attached.
b) See the image below: ↓