Question

Compute and simplify the difference quotient
f(a+h)-f(a)
---------
h
g(x)=2x^2+14x -13

Answer 1

this was already confusing enough without changing it to g(x)

Answer 2

replace \(g\) by \(f\) it makes no difference, it is all algabra

Answer 3

oh ok. so it's like: 2(a+h)^2+14(a)-13 over h?

Answer 4

which comes to...

Answer 5

no hold on

Answer 6

oh

Answer 7

you have to subtract first

Answer 8

well, first you have to expand, then you have to subtract

Answer 9

\(g(a+h)= 2(a+h)^2+14(a+h)-13\)

Answer 10

expand like 2a+2h.... and so on?

Answer 11

(2a+2h)^2?

Answer 12

for the first one

Answer 13

yeah it is
\[2(a^2+2ah+h^2)+14a+14h-13\]
\[=2a^2+4ah+2h^2+14a+14h-13\]

Answer 14

square first multiply by 2 second

Answer 15

oh ok, gotcha

Answer 16

from this you need to subtract \(2a^2+14a-13\) so everything without an \(h\) will be gone

Answer 17

then you can factor out an \(h\) from the remaining terms in the numerator, and cancel with the \(h\) in the denominator

Answer 18

you should be left with
\[4ah +14h+2h^2\] before you divide
after you divide you get
\[4a+14\] and next week you will be able to do this problem instantly in your head

Answer 19

ok, so I have 2a^2 +4ah + 2h^2 +14a +14h -13 and then I subtract 2a2+14a−13? how do I know to do that?

Answer 20

subtract

Answer 21

\(2a^2-2a^2=0\) \(14a-14a=0\) and \(-13-(-13)=0\)

Answer 22

in other words, every term without an \(h\) is gone

Answer 23

oh, now I get it

Answer 24

and that leaves me with 4ah + 2h^2 +14h

Answer 25

yes, and now divide by \(h\)

Answer 26

as you said lol

Answer 27

ok

Answer 28

4a+2h+14

Answer 29

which is one of my multiple choice answers...

Answer 30

well, I dunno if I'll be able to do that in my head instantly, but if I can work it out, then I'll be ok at least haha. Thanks @satellite73