- IF F OF X 
= 6 - 5X + 2X SQUARED,
WE WANT TO FIND F PRIME OF X 
AND F PRIME OF 2.
SO TO FIND 
THE DERIVATIVE FUNCTION,
WE'LL APPLY THE DERIVATIVE RULES 
GIVEN HERE BELOW.
SO F PRIME OF X WILL BE EQUAL 
TO THE DERIVATIVE OF 6
WHICH WOULD BE EQUAL TO 0,
THE DERIVATIVE OF ANY CONSTANT
IS ALWAYS 0 - 5 
x THE DERIVATIVE OF X
WHICH WOULD JUST BE 1 + 2 
x THE DERIVATIVE OF X SQUARED
WHICH WOULD BE 2 
x X TO THE POWER OF 2 - 1
OR X TO THE 1st, 
SO x 2X TO THE 1st.
LET'S GO AHEAD AND SIMPLIFY.
OUR DERIVATIVE FUNCTION IS EQUAL 
TO--THIS WOULD JUST BE -5 + 4X.
NOW THAT WE HAVE 
THE DERIVATIVE FUNCTION
WE CAN DETERMINE F PRIME OF 2
WHICH WOULD GIVE US THE SLOPE 
OF THE TANGENT LINE AT X = 2.
SO F PRIME OF 2 WOULD BE EQUAL 
TO -5 + 4 x 2.
THIS WOULD BE -5 + 8 OR 3.
SO WE HAVE 
OUR DERIVATIVE FUNCTION,
AND WE FOUND F PRIME OF 2.
BEFORE WE LOOK AT THE GRAPH 
OF THIS, THOUGH,
F PRIME OF 2 = 3 
TELLS US THAT WHEN X = 2
THE SLOPE OF THE TANGENT LINE 
WOULD BE 3.
LET'S GO AHEAD AND TAKE A MOMENT 
AND FIND THE POINT OF TANGENCY
OR THE POINT ON THE FUNCTION
WHERE THE SLOPE OF THE TANGENT 
LINE WOULD BE 3.
WE ALREADY KNOW 
THE X-COORDINATE IS GOING TO 2.
TO FIND THE Y-COORDINATE,
WE WOULD HAVE TO SUBSTITUTE 2 
BACK INTO THE ORIGINAL FUNCTION,
SO LET'S GO AHEAD AND DO THAT.
F OF 2 WOULD BE EQUAL 
TO 6 - 5 x (2 + 2) x 2 SQUARED,
SO THIS WOULD BE 6 - 10 
+ 8 WHICH WOULD BE 4.
SO THE POINT OF TANGENCY
WHERE THE TANGENT LINE HAS 
A SLOPE OF 3 WOULD BE (2,4).
SO LET'S GO AHEAD AND TAKE 
A LOOK AT THE GRAPH OF THIS.
HERE'S A GRAPH 
OF OUR QUADRATIC FUNCTION.
HERE'S THE POINT OF TANGENCY 
THAT WE FOUND, (2,4),
AND SINCE F PRIME OF 2 
WAS EQUAL TO 3
WE KNOW THE SLOPE 
OF THIS TANGENT LINE = 3.
I HOPE THIS WAS HELPFUL.
