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

Mar 16, 2016

$\text{ } y = \frac{1}{7} x - \frac{15}{7}$

#### Explanation:

The gradient of the line is found by differentiation

Let $\text{ } y = - 2 {x}^{3} - 3 {x}^{2} + 5 x - 2$ .......................(1)

Gradient $= \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {x}^{2} - 6 x + 5$

Thus the gradient of $f \left(x\right)$ at $x = 1$ is

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {\left(1\right)}^{2} - 6 \left(1\right) + 5 = - 7$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The gradient of the line that is normal to $f \left(x\right)$ is
$\text{ } \left(- 1\right) \times \frac{\mathrm{dx}}{\mathrm{dy}} = + \frac{1}{7}$

So we have

$\text{ } y = \frac{1}{7} x + c$......................(2)

We have a known point ${P}_{\text{(x,y)") ->P_("(1,y)}}$

We can determine the value of y by $f \left(x\right) \to f \left(1\right)$

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

$\text{ } y = - 2$

We now have ${P}_{\text{(x,y)") ->P_("(1,-2)}}$

So by substitution in equation (2)
$\text{ } - 2 = \frac{1}{7} \left(1\right) + c$......................(2)

$\text{ } c = - \frac{15}{7}$

Giving

$\text{ } y = \frac{1}{7} x - \frac{15}{7.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$