# What is the equation of the normal line of f(x)= x^2/(x^3− 5x + 1) at x = 5?

Nov 26, 2017

Equation of normal line is $y = 13.79 x - 68.7$

#### Explanation:

$f \left(x\right) = {x}^{2} / \left({x}^{3} - 5 x + 1\right) \therefore f \left(5\right) = {5}^{2} / \left({5}^{3} - 5 \cdot 5 + 1\right) \approx 0.25 \left(2 \mathrm{dp}\right)$

Point is at $\left({x}_{1} , {y}_{1}\right) = \left(5 , 0.25\right)$ . Slope is $m = f ' \left(x\right)$

$f ' \left(x\right) = \frac{2 x \cdot \left({x}^{3} - 5 x + 1\right) - {x}^{2} \left(3 {x}^{2} - 5\right)}{{x}^{3} - 5 x + 1} ^ 2$

$f ' \left(x\right) = m = \frac{25 \cdot \left({5}^{3} - 5 \cdot 5 + 1\right) - {5}^{2} \left(3 \cdot {5}^{2} - 5\right)}{{5}^{3} - 5 \cdot 5 + 1} ^ 2$ or

$m \approx - 0.07$. Slope of normal is ${m}_{1} = - \frac{1}{-} 0.07 \approx 13.79 \left(2 \mathrm{dp}\right)$

Equation of normal line is $\left(y - {y}_{1}\right) = {m}_{1} \left(x - {x}_{1}\right)$ or

$\left(y - 0.25\right) = 13.79 \left(x - 5\right) \mathmr{and} y = 13.79 x - 68.7$ [Ans]