# What is the equation of the line normal to f(x)=9x^2 +2x - 4  at x=-1?

Jun 30, 2016

$y = + \frac{1}{16} x + \frac{49}{16}$

#### Explanation:

Let the given point be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 1 , {y}_{1}\right)$

Differentiating each term of $f \left(x\right)$

$9 {x}^{2} \to 2 \times 9 {x}^{2 - 1} = 18 x$

$2 x \to 2$

$- 4 \to 0$

So $\frac{\mathrm{dy}}{\mathrm{dx}} \to {f}^{'} \left(x\right) = 18 x + 2$

$\implies {f}^{'} \left(- 1\right) = 18 \left(- 1\right) + 2 = - 16$
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Gradient of the line normal to $f \left(x\right) \text{ is } \left(- 1\right) \times \frac{1}{- 16}$ giving:

$\text{ } \textcolor{b l u e}{y = + \frac{1}{16} x + c}$ ................................Eqn(1)

This line passes through the point $\left(x , y\right) \to \left(- 1 , {y}_{1}\right)$

Substitute ${x}_{1} = - 1$ into $f \left(x\right) \to f \left(- 1\right) \text{ " ->" } {y}_{1} = 9 {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 4$

${y}_{1} = 9 - 2 - 4 = + 3$

Thus ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 1 , 3\right)$
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So Eqn(1)$\to {P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to {y}_{1} = \frac{1}{16} {x}_{1} + c \text{ }$ becomes

$\implies {P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to 3 = \frac{1}{16} \left(- 1\right) + c$

Thus $c = + 3 \frac{1}{16} \to \frac{49}{16}$
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The equation of the line is:

$y = + \frac{1}{16} x + \frac{49}{16}$ 