Calculus question, what exactly is x=a??

Question

Consider the function
y =x^2-3x+4
(a) Determine the slope of the tangent to the curve at x = a.
(b) Show that if the straight line y = mx + k is a tangent to the curve then
(m + 3)^2 = 4(k − 4).

Answer 1

a) Slope of thetangent to the curve  at x=a is equal to f'(a)=2a-3,  

[ f'(x)=2x-3]

b) Since  y=mx+k is  a tangent to y=x^2-3x+k, then m=2x-3, and

x=(1/2)(m+3)

At the tangent point it should be 

x²-3x+4=mx+k

Substitite x=(1/2)(m+3)

(1/4)(m+3)²-(3/2)(m+3)+4=(m/2)(m+3)+k

(1/4)(m+3)²-(m+3)(m+3)/2=k-4

(1/4)(m+3)²=(k-4)  (multiply both sides by 4)

(m+3)²=4(k-4),

what is required to show.

Answer 2

No fear! The question is referring to one of the two different quotients used to find a derivative (until, that is, they show you the major shortcuts...). Your difference quotient is:

 

Lim     f(x) - f(a)

x→a      x - a

 

You already have a formula using x:

f(X) = X² -3X + 4

but you must also plug in "a" for X to get...

f(A) = A² - 3A + 4

They go into the difference quotient to make:

f’(X) = Lim      (X²-3X+4) - (A²-3A+4)

           x→A              X - A

This step varies depending on the problem, but basically, simplify and/or factor as needed:

f’(X) = Lim       X²-3X+4-A²+3A-4

           X→A             X  -  A

= Lim       X²-3X-A²+3A

   X→A             X  -  A             And now, the fun part! You need factoring by grouping! Sort of...

= Lim       X²-A² -3X+3A

   X→A            X  -  A

= Lim       (X-A)(X+A) -3(X-A)

   X→A             X  -  A

= Lim        (X-A)(X+A-3)          I factored out the "X-A", leaving everything else behind.

   X→A             X  -  A              Now, the (X-A) cancels, leaving...

= Lim        X+A-3

   X→A

= A+A-3 = 2A - 3

Or, in terms of X...

= 2X - 3

b) I will leave this for you with a hint that we just found the slope, often written as "m". If you replace the m in their example with your slope from part I, you will get their tangent equation.