Question

compute the difference quotient of the given function

Z(y)=3-8y-y^2

Answer 1

same as your previous problem.... you'll need to compute: \(\large Z(y+h) \) for
\(\large Z(y)=3-8y-y^2 \)

then simplify the expression: \(\large \frac{Z(y+h)-Z(y)}{h} \)

Answer 2

ok

Answer 3

would I plug this one in the same way as the other one?

Answer 4

@dpaInc

Answer 5

yeah the procedure is the same.... just a different function....
\(\large Z(y+h)=3-8(y+h)-(y+h)^2 \)
simplify that...

Answer 6

expand and simplify that expression then put that into the difference quotient....

sorry.... i gotta leave.... someone should be by to help u.... do you want me to tag someone i recommend?

Answer 7

@Callisto , please help...

Answer 8

ok thanks @dpalnc

Answer 9

Hello, what's going on here?! May I help?

Answer 10

yesss... longer problems are harder for me to work out especially if I haven't yet experienced calculus yet...

Answer 11

@callisto

Answer 12

Can you work out what Z(y+h) is?

Answer 13

=3-8

Answer 14

Huh?!
Z(y)=3-8y-y^2
Z(y+h) => replace y by y+h in Z(y)=3-8y-y^2..
So, what do you get?

Answer 15

I don't know.... i"ll try to figure it out on my own thanks

Answer 16

Z(y+h) = 3-8(y+h) - (y+h)^2..
Since you have to replace y by y+h in Z(y)

Then, you have to expand it. What do you get from the expansion?