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

Aug 19, 2017

$y = - 5 x + 6$

#### Explanation:

•color(white)(x)m_(color(red)"tangent")=f'(x)" at x = a"

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

$g \left(x\right) = - {x}^{2} - x + 3 \Rightarrow g ' \left(x\right) = - 2 x - 1$

$h \left(x\right) = 2 x - 1 \Rightarrow h ' \left(x\right) = 2$

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

$\Rightarrow f ' \left(1\right) = \frac{- 3 - 2}{1} = - 5$

$\Rightarrow f \left(1\right) = 1 \to \left(1 , 1\right) \leftarrow \text{ tangent point}$

$\Rightarrow y - 1 = - 5 \left(x - 1\right)$

$\Rightarrow y = - 5 x + 6 \leftarrow \text{ equation of tangent}$