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

##### 2 Answers

# y = -1/6x+5/6 #

#### Explanation:

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

We have:

# f(x) = (x+2)^2-2x^2+2 #

# " " = x^2+4x+4-2x^2+2 #

# " " = -x^2+4x+6 #

Differentiating wrt

# dy/(dx) = -2x+4 #

So when

# y = -1-4+6=1# and#dy/dx=2+4=6#

So the tangent/normal passes through

# y - 1 = -1/6(x+1) #

# y - 1 = -1/6x-1/6 #

# y = -1/6x+5/6 #

We can verify this graphically:

#### Explanation:

One way to think of the derivative is as a "slope-finding-formula". The result of a derivative is a function, such that when you plug in values for **tangent** line.

The **normal** line is just the perpendicular line through the tangent line, *at the same point*. Recall from algebra that if you have a slope

Perpendicular slope:

First, determine the exact point where your normal line will pass through the given curve. This happens at

The point that our normal line will go through is

Second, find the derivative, or slope-finding formula. It may help to fully expand out the original function first.

Then find the derivative

The slope of the tangent line at

Therefore, the slope of the perpendicular **normal** line is

Now you have all the parts you need to determine the formula for this normal line. You have a slope,