Question

Tangent lines Determine an equation of the line tangent to the graph of y=(x²−1)² / x³−6x−1 at the point (0,−1).

Accepted Answer

Step 1: Identify the function and the point of tangency. The function is $$ y = \frac{(x^2 - 1)^2}{x^3 - 6x - 1} $$ and the point of tangency is $$(0, -1). $$Step 2: Find the derivative of the function $$ y $$ with respect to $$ x  to $$determine the slope of the tangent line. Use the quotient rule: $$ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} $$, where $$ u = (x^2 - 1)^2 $$ and $$ v = x^3 - 6x - 1 . $$Step 3: Calculate $$ u' $$ and $$ v' . $$For $$ u = (x^2 - 1)^2 $$, use the chain rule: $$ u' = 2(x^2 - 1) \cdot 2x = 4x(x^2 - 1) . $$For $$ v = x^3 - 6x - 1 $$, $$ v' = 3x^2 - 6 . $$Step 4: Substitute $$ u, u', v, $$ and $$ v' $$ into the quotient rule formula to find $$ y' . $$Simplify the expression to find the derivative. Step 5: Evaluate the derivative $$ y'  at$$ $$ x = 0  to $$find the slope of the tangent line at the point $$(0, -1). $$Use the point-slope form of a line, $$ y - y_1 = m(x - x_1) $$, where $$ m  is $$the slope and $$(x_1, y_1) is $$the point of tangency, to write the equation of the tangent line.

Tangent lines Determine an equation of the line tangent to the graph of y=(x²−1)² / x³−6x−1 at the point (0,−1).

Verified video answer for a similar problem:

Key Concepts

Tangent Line

Derivative

Point-Slope Form