# How do I find the equation of the normal line given a point on the curve?

The **normal line** to a curve at a particular point is the line through that point and *perpendicular* to the tangent.

A simple trick to remembering how to find the normal gradient, ** n**, is that the slope of any line perpendicular to a line that has a gradient,

**is just the negative reciprocal,**

*m,***.**

*-1/m*
__Example:__

**Find the normal gradient to the curve **y=2x^{3 }+3x+7 **at the point (1,1).**

So firstly, let’s recap how to calculate the gradient of the tangent line:

By differentiating y=2x^{3 }+3x+7 , we find

dy/dx = 6x^{2} +3

Then, by substituting in our point, at x=1 we yield **dy/dx=9**. This is the tangent gradient of the curve (m=9).

Finally we substitute this into our formula for calculating the normal gradient n=-1/m.

Therefore **n=-1/9**.

Now, let’s try another example which demonstrates how we use the normal gradient to find an equation for the normal line.

We will use the formula (y-y_{0}) = n(x-x_{0}), where (x_{0},y_{0}) is a given point.

__Example:__

**Consider a curve **y=x^{5}+3x^{2} +2. **Find the equation of the normal to the curve at the point (-1,2). Leave your answer in the form y=mx+c.**

By differentiating the curve, we have dy/dx = 5x^{4 }+6x.

To find the gradient of the tangent line we substitute in x=-1, which yields

dy/dx = 5(-1)^{4} +6(-1)

= 5-6

=-1 = m

Therefore, we know that the normal gradient is n=-1/m

So **n=1**

Finally, we substitute this into our formula for the normal line (y-y_{0}) = n(x-x_{0}):

In our example, (x_{0}, y_{0}) = (-1,2)

So y-2 = 1(x+1)

And my rearranging, we find **y = x+3.**