The normal line to a function at a point is the line perpendicular to its tangent line. The normal line to a function has a slope which is the negative reciprocal of its derivative. 

For our function , the derivative is .

We now may plug in our point of , , However, since we seek the negative reciprocal it now becomes

Our normal line will have the form:

First, we take the derivative of the line, which gives us the slope of the curve at any point

Then, we evaluate the slope at the point  to obtain an actual value

To solve the problem, we simply take the derivative of the function and evaluate it at the specified point. The derivative of a line is the slope of the curve, so we do the following

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

The derivative was found using the following rule:

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is 

However, the negative reciprocal of this - the slope of the tangent line - is

We now can write the equation of the line using point-slope form:

To determine the equation of the line, we must first find its slope, which is parallel to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

which was found using the following rules:

, , 

Now, we find the slope of the tangent line at the given point by plugging this point into the first derivative function:

Now, we use point-slope form to find the equation of the line:

To determine the equation of the line, we must first find its slope, which is parallel to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

The following rule was used to find the derivative:

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

We now can write the equation of the line using point-slope form:

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

which was found using the following rules:

, 

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

However, the line we want has a slope perpendicular to this, so we take the negative reciprocal:

We now can write the equation of the line using point-slope form:

Find the slope of the line tangent to the curve of f(x) when 

To find the slope of a tangent line, we need to find our first derivative.

To find our derivative, we need to recall two rules.

And

Using these two rules, we can find the derivative of f(x).

Our first term can be derived using our first rule. The derivative of e to the x is just e to the x.

This means that our first term will remain 16e to x.

For our other three terms, we follow the second rule. We will decrease each term's exponent by 1, and then multiply the coefficient by the old exponent.

Notice that the 13 will drop out. It is a constant term, and as such when we multiply it by it's original exponent (0) it wil be reduced to zero as well.

Clean up the above to get:

Now, we are almost there. We need to find the slope when x=-2. To do so, plug in -2 for x and solve.

So, our answer is -2350

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

The first derivative was found using the following rules:

, , 

Evaluating the derivative at the given point, we get

Because the slope of the line we want is perpendicular to this line, its slope is the negative reciprocal, or

We can now find the equation of the line using point-slope form:

The slope of any point on this function can be determined by plugging the point's x-value into the , the first derivative of .