To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

Taking the first derivative using the Power Rule  we get the following.

Substituting in 6 for b and solving we get:

.

So our answer is 320160

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

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

Applying this rule we get:

Therefore, the slope we are looking for is 454.

To find the slope of the line at that point, find the derivative of f(x) and plug in that point. 

Remember that the derivative of  and the derivative of   

Now plug in  

The first step here is to integrate  in order to get .

Here the problem tells us that the integration constant , so

Plug in  here

The slope of a curve at any point is equal to the derivative of the curve at that point.

Remembering that the derivative of  and using the power rule on the second term we find the derivative to be:

.

Pluggin in  we find that the slope is .

Find the line tangent to  at .

First, we find :

Next, we find the derivative:

Therefore, the slope at  is:

.

Using point-slope form, we can write the tangent line:

Simplifying this gives us:

The other point of the triangle must be at  as it must be equidistant from the other two points of the triangle. Since all points on the y-axis are  units away from the other points in the  direction, the third point must be equidistant in the  direction from both  and . The distance between these points is , so the third point must have a y-value of . The third point is now at  so the slope of the line from  to  is as follows.

We must take the derivative of the function using the chain rule yielding .

The chain rule is .

Also remember that the derivative of  is .

Applying these rules we get the following.

Plugging in the value for  we get  which is .

A line is tangent to a point on a curve when its slope is equivalent to the slope of the curve at the point of intersection. Therefore to solve this equation, the slope of the curve at  must be found.  

The slope of a graph at any point can be found by taking the first derivative.  To take the derivative of this equation, we must use the power rule,  

.  

We also must remember that the derivative of an constant is 0.  By taking the first derivative of the graph equation, we obtain the slope equation 

.  

Plugging in , we find the slope at that point is 13, therefore any line tangent to the curve at that point must have a slope of 13.