# Ex 2: The Difference Quotient (Quadratic Function)

– WELCOME TO A SECOND EXAMPLE OF DETERMINING THE DIFFERENCE
QUOTIENT FOR A GIVEN FUNCTION. IN THIS EXAMPLE, THE GIVEN
FUNCTION IS A QUADRATIC FUNCTION AND AS MENTIONED IN EXAMPLE 1, THIS DIFFERENCE QUOTIENT
OR THIS QUOTIENT HERE IS A VERY IMPORTANT QUOTIENT
IN THE STUDY OF CALCULUS, BUT FOR THIS EXAMPLE WE’RE JUST
WORKING ON OUR ALGEBRA SKILLS TO SIMPLIFY THIS QUOTIENT. SO TO START WE WANT TO FIND
F OF THE QUANTITY X + H, WHICH MEANS THE QUANTITY X + H IS THE INPUT INTO
THE GIVEN FUNCTION. SO WHEREVER WE SEE X, WE’LL REPLACE X
WITH THE QUANTITY X + H. SO IF (F OF X)
=X SQUARED – (3X + 4), F OF THE QUANTITY X + H
WOULD BE, INSTEAD OF X SQUARED,
(X + H SQUARED), AND THEN INSTEAD OF – 3X WE’LL HAVE – 3
x (THE QUANTITY X + H) + 4. SO ALL OF THIS IS
F OF THE QUANTITY X + H. LET’S PUT THIS IN PARENTHESES
TO KEEP THINGS CLEAR AND NOW WE STILL HAVE
TO SUBTRACT (F OF X) SO WE’LL HAVE – (THE QUANTITY X
SQUARED) – (3X + 4). ALL THIS IS DIVIDED BY H. NOW WE’RE GOING TO MULTIPLY OUT
(F OF THE QUANTITY X + H) SO WE’LL START BY SQUARING
(THE QUANTITY X + H). THERE’S NO SHORTCUTS HERE. IF WE WANT TO SQUARE
(THE QUANTITY X + H) WE’LL HAVE TWO FACTORS
OF (X + H). WHEN MULTIPLYING THESE BINOMIALS
WE’LL HAVE 4 PRODUCTS, ONE, TWO, THREE AND FOUR. SO X x X=X SQUARED.
X x H=+HX. H x X IS ANOTHER +HX,
AND THEN FINALLY +H SQUARED. SO WE’LL HAVE X SQUARED PLUS,
WE HAVE 1HX + 1HX, THAT’S 2HX + H SQUARED. WHICH MEANS
(F OF THE QUANTITY X + H) WOULD BE (X SQUARED + 2HX)
+ H SQUARED, AND NOW WE’LL DISTRIBUTE -3, SO WE HAVE (-3X – 3H) + 4
AND WE STILL HAVE (-F OF X). SO – (THE QUANTITY X SQUARED
– 3X) + 4. THIS IS STILL DIVIDED BY H. NOW WE’LL CLEAR THE PARENTHESES
IN THE NUMERATOR AND COMBINE LIKE TERMS. SO IF IT’S HELPFUL WE CAN THINK
OF DISTRIBUTING A +1 HERE WHICH WON’T CHANGE ANYTHING,
BUT BECAUSE OF THE SUBTRACTION, WE CAN THINK OF DISTRIBUTING
A -1 IF THAT’S HELPFUL. BUT FOR THE FIRST PART
WE WOULD JUST HAVE (X SQUARED + 2HX) +
(H SQUARED – 3X) – (3H + 4). NOTICE NOTHING CHANGED
BUT FOR THE SECOND PART IF WE DISTRIBUTE -1 OR SUBTRACT
THIS ENTIRE FUNCTION, IT’S GOING TO CHANGE THE SIGN
OF EACH TERM, SO WE’LL HAVE – X SQUARED
AND THEN (-1 x -3X)=+3X AND THEN -1 x +4 WOULD BE -4. NOW LOOKING AT THE NUMERATOR NOTICE HOW WE HAVE
X SQUARED – X SQUARED, THAT’S 0. WE ALSO HAVE -3X + 3X,
THAT’S 0 AND THEN WE ALSO HAVE 4 – 4,
THAT’S 0. SO WE’RE LEFT WITH (2HX + H SQUARED)
– 3H DIVIDED BY H BUT WE’RE NOT DONE YET.
THIS DOES SIMPLIFY FURTHER. NOTICE EACH TERM IN
THE NUMERATOR CONTAINS A COMMON FACTOR OF H. SO WE’RE GOING TO FACTOR OUT THE COMMON FACTOR OF H
FROM THE NUMERATOR. THAT WILL LEAVE US WITH (H x THE QUANTITY 2X) + (H – 3)
STILL DIVIDED BY H, BUT NOW WE HAVE H/H,
THAT SIMPLIFIES TO 1, SO OUR SIMPLIFIED DIFFERENCE
QUOTIENT IS JUST (2X + H) – 3. OKAY. I HOPE YOU FOUND THIS