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hint:
the slope m of your secan t, is:
\[m = \frac{{{\delta _y}}}{{{\delta _x}}}\]

furthermore, we can write this:
\[\Large 4 + {\delta _y} = {\left( {2 + {\delta _x}} \right)^2}\]

you then solve for delta y? As values for delta x are already given?

hmm

|dw:1438251569667:dw|

delta x= E means delta x= infinity?

sorry, epsilon

I think you want to finish part \(a\) first
let \(\delta_x=0.01\) and find the slope of secant

knew what you meant :)

\[4+\delta y=4+(\delta x)^2+4 \delta x\] then subtracted 4 on both sides to isolate \[\delta y\]

That looks good!

then got \[( \delta x)^2+4 \delta x\] then sub in my x value for a)

For part \(a\), below works too :
\[m= \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{(2+0.01)^2-2^2}{0.01}\]

thanks

|dw:1438252734043:dw|

Yes, plugin \(\delta_x = \epsilon\) and simplify the secant expression as much as you can

great, thank you so much. So much clearer now :)

As simplified as I can? is that what you mean?

Yes, but I have simplified that for you already

when using \[ \frac{ \delta y }{ \delta x }\] I get a different result than y2-y1/x2-x1

both should give you same answer, show me ur work

and with \[\frac{ y2 -y1}{ x2-x1 }=\frac{ (2+0.01)^2-2^2}{ 0.01 } =4.01\]

looks fine, whats wrong with that ?

notice that slope of secant equals \(\dfrac{\delta y}{\delta x}\), not just \(\delta y\)

ok, so my first one i still need to divide by delta x. then they will be the same. :)

Yep! Easy, slope = (change in y)/(change in x)

thats the highschool definition of slope, and it still works! :)

Awesome, thank you so much. Lifesaver!

Great, thanks!

np