SOLUTION: Let {{{ f(x) = x^3 + 11 }}} and {{{ a = 3 }}} Find and simplify the quotient: {{{(f(x)-f(a))/(x - a)}}}= Then find the slope {{{ M_a }}} of the line tangent to the grap

First of all, if you use triple brackets improperly, 
it messes everything up.  So don't use them next time
if you you don't understand the triple bracket notation 
program used on this site  I rewrote your problem 
correctly in the triple bracket program.

Let  and  
Find and simplify the quotient:



Then find the slope  of the line tangent to the graph 
of f at the point (a,f(a))  

     

---------------------

Since they tell you that a=3, rewrite the problem using 3 for a
and it will make more sense:


Find and simplify the quotient:



Then find the slope  of the line tangent to the graph 
of f at the point (3,f(3))  



------------------------------

Now since you know what f(x) is, you can find f(3),


So now rewrite the problem with 38 for f(3)

--------------------------------

Find and simplify the quotient:



Then find the slope  of the line tangent to the graph 
of f at the point (3,38)  



--------------------------------

We are trying to find the slope of a line tangent to the graph of

 at the point (3,38)

The slope formula  when applied to the

two points , which is (3,38) and  gives 



This is the slope of the line that goes through the two points
(3,38) and the variable point 

As the variable point  gets closer and
closer to the fixed point  which is the
point (3,38), the line through them gets closer and closer to being 
a tangent line. We would like to make x equal to 3, for that would 
cause the line to become tangent but we can't yet.

We cannot yet choose x=3 in the slope formula


 because we get:



which gives 0 in the denominator.

So you see we cannot substitute x=3 yet, because the 
denominator will be zero.

Next we substitute  for  and
 for  in





and simplify:





Factor the numerator as the difference of two cubes:



Now we can cancel the (x-3)'s leaving



Now we CAN let x=3 because we no longer have a 
denominator to be 0.

So when we substitute x=3 we get





That's the slope of a line tangent to the graph of

 at the point (3,38)

So that slope 

Edwin