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

Aug 20, 2016

$y = \frac{1}{7} x + 24 \frac{6}{7}$

#### Explanation:

Given:$\text{ } f \left(x\right) = 4 {x}^{2} - 3 x - 2$.......................Equation(1)

$\implies {f}^{'} \left(x\right) = 8 x - 3$

So the gradient of the normal will be $\frac{1}{f} ^ ' \left(x\right)$

Thus we have $y = \frac{1}{f} ^ ' \left(x\right) x + c$...........................Equation(2)
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$\textcolor{b l u e}{\text{Determine the point on the curve at } x = 3}$

$f \left(3\right) = 4 {\left(3\right)}^{2} - 3 \left(3\right) - 2 = 25$

Thus for the point $\left(x , y\right) \to \left(3 , 25\right)$ equation(2) becomes:

$25 = \frac{1}{8 \left(3\right) - 3} \left(3\right) + c$

$c = 25 - \frac{1}{7} = 24 \frac{6}{7}$

By substitution Equation(2) becomes:

$y = \frac{1}{7} x + 24 \frac{6}{7}$