# What is the equation of the line normal to f(x)=-4x^2 -5x +1  at x=-3?

Feb 20, 2016

$y = - \frac{x}{19} - \frac{383}{19}$

#### Explanation:

Let $y = - 4 {x}^{2} - 5 x + 1$

$y ' = - 8 x - 5$

This is the gradient of the tangent $m$.

At $x = - 3$ this becomes:

$m = \left(- 8 \times - 3\right) - 5 = 19$

If the gradient of the normal is $m '$ then:

$m ' m = - 1$

$\therefore m ' = - \frac{1}{19}$

To get the value of $y$ at $x = 3 \Rightarrow$

$y = - 4 \times {\left(- 3\right)}^{2} - \left(5 \times - 3\right) + 1$

$y = - 20$

The equation of the normal line is of the form:

$y = m ' x + c$

$\therefore - 20 = \left(- \frac{1}{19} \times - 3\right) + c$

$\therefore c = - 20 - \frac{3}{19} = - \frac{380}{19} - \frac{3}{19} = - \frac{383}{19}$

So the equation is:

$y = - \frac{x}{19} - \frac{383}{19}$

This is shown here:

graph{(-4x^2-5x+1-y)(-x/19-383/19-y)=0 [-80, 80, -40, 40]}