Find the equation of the tangent line to the graph of f (x) = x3 + 4х - 1) (x - 2) at the given point, (1, -4) . Write your answer in slope-intercept form, y = mx + b.

Answer 1

Let f be the function defined by

f(x) = (x³ + 4x - 1) / (x - 2)

for all values of x except x = 2.

Using the Quotient Rule for derivatives, we find that the derivative of the function f is given by:

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

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

= (3x³ + 4x - 6x² - 8 - x³ - 4x + 1) / (x - 2)²

= (2x³ - 6x² - 7) / (x - 2)²

The derivative of the function f at x = 1 is given by:

f'(1) = (2 * 1³ - 6 * 1² - 7) / (1 - 2)²

= (2 * 1 - 6 * 1 - 7) / (- 1)²

= (2 - 6 - 7) / 1

= -11 / 1

= -11

Thus, we know that the tangent line at x = 1

has a slope of -11 and goes through the point (1, -4).

An equation of that tangent line in point slope form is given by:

y - (-4) = -11(x - 1)

We just have to get this equation to slope intercept form.

y - (-4) = -11(x - 1)

⇒ y + 4 = -11x + 11

⇒ y + 4 - 4 = -11x + 11 - 4

⇒ y = -11x + 7