# What is the equation of the line tangent to  f(x)=(-x^2-1)/(x^2+4)  at  x=-3 ?

Jun 4, 2018

The tangent line at that point is $y + \frac{10}{13} = \frac{126}{169} \left(x + 3\right)$.

#### Explanation:

We need to find the slope of the tangent line at this point, $x = - 3$. To do so, we find $f ' \left(x\right)$ and plug -3 in:

$f ' \left(x\right) = \frac{- 2 x \left({x}^{2} + 4\right) - 2 x \left({x}^{2} - 1\right)}{{x}^{2} + 4} ^ 2 = \frac{- 2 {x}^{3} - 8 x - 2 {x}^{3} + 2 x}{{x}^{+} 4} ^ 2 = \frac{- 4 {x}^{3} - 6 x}{{x}^{2} + 4} ^ 2$

$f ' \left(- 3\right) = \frac{\left(- 4 \cdot - 27\right) - \left(6 \cdot - 3\right)}{9 + 4} ^ 2 = \frac{126}{169}$

We now find the $y$-value at $x = - 3$:

$f \left(- 3\right) = \frac{- 9 - 1}{9 + 4} = - \frac{10}{13}$

Using point-slope form, we have the equation of the tangent line as

$y + \frac{10}{13} = \frac{126}{169} \left(x + 3\right)$