Equation of a tangent line

Question

Answer 1

Evaluate the derivative of f(x) at x=1. That will give you the slope m of the tangent line at x=1.

f'(x) = [-3x²(x + x²) - (4 - x³)(1 + 2x)] / (x + x²)²

m = f'(1) = ?

Evaluate f(1). That will give you the tangent point (1,f(1)). Then finish by writing the equation of the line in point slope form.

y - y₁ = m(x - x₁)

x₁ = 1

y₁ = f(1)

m = f'(1)

You have everything you need, just plug in the numbers and evaluate.

Answer 2

Use quotient rule to evaluate the derivative. (If you don't know the quotient rule, say that in a comment and I will try to help further).

The slope is -9/4. (evaluate the derivative at x=1)

The tangent line is -9/4 = {[y-(3/2)]/(x-1)} and you can "massage" this around into any desired form. Remember the tangent goes through the point on the graph where x=1).