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

$x + 7 y = - 6$

#### Explanation:

We need to obtain first the slope m which is negative reciprocal of the slope of the tangent line at x=1 on the curve $f \left(x\right) = 3 {x}^{2} + x - 5$

Compute the first derivative $f ' \left(x\right)$ and $m = f ' \left(1\right)$

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} + x - 5\right) = 6 x + 1$

$f ' \left(x\right) = 6 x + 1$

at $x = 1$

$f ' \left(1\right) = 6 \left(1\right) + 1 = 7$

Compute $m$

$m = - \frac{1}{f ' \left(1\right)} = - \frac{1}{7}$

Determine the ordinate for the given abscissa $x = 1$ using the original equation $f \left(x\right) = 3 {x}^{2} + x - 5$

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

The point is at $\left(1 , - 1\right)$

Determine the Normal Line now using Point-Slope Form

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - - 1 = - \frac{1}{7} \left(x - 1\right)$

$7 \left(y + 1\right) = - x + 1$

$7 y + 7 = - x + 1$
$x + 7 y = - 6$

See the graph of $f \left(x\right) = 3 {x}^{2} + x - 5$ and $x + 7 y = - 6$

graph{(y-3x^2-x+5)(x+7y+6)=0[-22,22,-10,10]}