This article is only useful if you know what tangets are. Be able to find gradients of tangents and be able to form general formulas for tangents. This is just an expansion from Gradient of a Tangent to a curve and without any knowledge of tangents be ready for huge headaches.

You know that a tangent to a curve is a straight line which touches the curve at a certain point. Well...

The normal to a curve is the straight line perpendicular to the tangent at the same point as the tangent attachment to the curve. In the following illustration the normal to the curve is the blue line while the orange line is the tangent and the red line the curve.

If the gradient of the tangent is m

...then the gradient of the normal is -1/m

For instance;

if the gradient of a tangent is 4 then the gradient of the normal is -1/4

It is as simple as that, this is how you find the normal to the curve at a certain point but you first need to differentiate to find the gradient first. In most cases when working with tangents and normals there will be a need to find the equations of the tangents and normals at a certain point. In the following examples we shall be exploring just that. To find the equations of either the tangent of the normal we need to know the gradient of the curve and the point of the line then we use the following equation;_{1}= m(x - x

_{1})

## Finding the equation of the tangent

In this section we shall look at how to find the equation of a tangent.^{2}+ 4 at the point where x = 1.

^{2}+ 4, you would get the following curve on the graph. [IMAGE]

We can also see the tagent (which is the line touching the curve at that point)

[IMAGE]^{2 - 1}+ 0

^{2}+ 4

_{1}= m(x - x

_{1}) to find the equation of the tangent to the point where x = 1.

_{1}= m(x - x

_{1})

_{1}= m(x - x

_{1})

^{2}+ 4 at the point where x = 1 is given by; y = 6x + 1.

## Finding the equation of a normal

_{1}= m(x - x

_{1})

^{2 - 1}- 3x

^{1-1}- 0

_{1}= m(x - x

_{1}) to find the equation of the normal to the point where x = 3.

_{1}= m(x - x

_{1})

_{1}= m(x - x

_{1})

This article is only useful if you know what tangets are. Be able to find gradients of tangents and be able to form general formulas for tangents. This is just an expansion from Gradient of a Tangent to a curve and without any knowledge of tangents be ready for huge headaches.

You know that a tangent to a curve is a straight line which touches the curve at a certain point. Well...

The normal to a curve is the straight line perpendicular to the tangent at the same point as the tangent attachment to the curve. In the following illustration the normal to the curve is the blue line while the orange line is the tangent and the red line the curve.

If the gradient of the tangent is m

...then the gradient of the normal is -1/m

For instance;

if the gradient of a tangent is 4 then the gradient of the normal is -1/4

It is as simple as that, this is how you find the normal to the curve at a certain point but you first need to differentiate to find the gradient first. In most cases when working with tangents and normals there will be a need to find the equations of the tangents and normals at a certain point. In the following examples we shall be exploring just that. To find the equations of either the tangent of the normal we need to know the gradient of the curve and the point of the line then we use the following equation;_{1}= m(x - x

_{1})

## Finding the equation of the tangent

In this section we shall look at how to find the equation of a tangent.^{2}+ 4 at the point where x = 1.

^{2}+ 4, you would get the following curve on the graph. [IMAGE]

We can also see the tagent (which is the line touching the curve at that point)

[IMAGE]^{2 - 1}+ 0

^{2}+ 4

_{1}= m(x - x

_{1}) to find the equation of the tangent to the point where x = 1.

_{1}= m(x - x

_{1})

_{1}= m(x - x

_{1})

^{2}+ 4 at the point where x = 1 is given by; y = 6x + 1.

## Finding the equation of a normal

_{1}= m(x - x

_{1})

^{2 - 1}- 3x

^{1-1}- 0

_{1}= m(x - x

_{1}) to find the equation of the normal to the point where x = 3.

_{1}= m(x - x

_{1})

_{1}= m(x - x

_{1})