Finding the equation of the tangent (in slope-intercept form) at a particular point?

Question

One of the practice problems in my Calculus book is as follows: The graph of y=$8/(x^2-4)$ is called the Witch of Agnesi.

(a) Find y'

d/dx (u/v) = (v du/dx - u dv/dx)/($v^2$)
$=((x^2+4)(0)-(8)(2x))/((x^2+4)^2)$
$=(-16x)/((x^2+4)^2)$

Assuming I've gotten (a) correct, I then am asked (b) Find the equation of the tangent (in slope-intercept form) at the point (2,1).

Can anyone help me get rolling on this one?

Answer 1

Here's a method to get you rolling:

The derivative of $f(x)$ is the slope of the tangent line at the point $(x, f(x))$. You already found the derivative to be $-\frac{16x}{(x^2+4)^2}$, so this is the slope of the tangent line at any point $(x, f(x))$.

To find the equation of the tangent line at $(2, 1)$, you first need to find the slope of the tangent line at $(2, 1)$, let's call it $m$. After you find $m$, you then plug $m$ and the point $(2, 1)$ into the point-slope formula and solve for $y$ $$ y - y_1 = m(x - x_1).$$

Answer 2

Undefined. Using your equation given and plugging $x=2$ gives $\frac80$ or undefined. The proper equation for the witch of agnesi is $\frac8{x^2+4}$.