# How do you find the equation of the tangent and normal line to the curve #y=6-x^2# at (2,2)?

##### 1 Answer

**Tangent:**

# y = -4x+10 #

**Normal:**

# y = 1/4x + 3/2 #

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is

We have:

# y = 6-x^2 #

First let us check that

# x=2 => y=6-4=2 #

Then differentiating wrt

# dy/dx = -2x #

When

So the tangent passes through

**Tangent:**

# y - 2 = -4(x-2) #

# :. y - 2 = -4x+8 #

# :. y = -4x+10 #

**Normal:**

# y - 2 = 1/4(x-2) #

# :. y - 2 = 1/4x-1/2 #

# :. y = 1/4x + 3/2 #

We can confirm this solution is correct graphically: