Let f(x)=2x2+4x+6. Then according to the definition of derivative f′(−8) is the limit as x tends to -8 of the expression.....?

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Let f(x)=2x2+4x+6. Then according to the definition of derivative f′(−8) is the limit as x tends to -8 of the expression.....?

Mathematics
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Oh hmm.. this is the other definition of the derivative ok.. lemme try to remember...
\[\large\rm \lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]I believe it looks like that. You have a secant line between the points (x,f(x)) and (a,f(a)) and you let x approach a to get yoor tangent line.
So umm

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\[\large\rm f(x)=2x^2+4x+6\]The derivative definition evaluated at a=-8 looks like: \[\large\rm \lim_{x\to a}\frac{f(x)-f(-8)}{x--8}\]So just plug in all the pieces! :)
\[\large\rm f(-8)=?\]Do you know how to figure this out?
yeah i have the answer, it's -28 but every time i plug the expression in it says it's wrong
\[\large\rm \lim_{x\to a}\frac{\color{royalblue}{f(x)}-\color{orangered}{f(-8)}}{x--8}=\lim_{x\to a}\frac{\color{royalblue}{2x^2+4x+6}-(\color{orangered}{2(-8)^2+4(-8)+6})}{x-(-8)}\]Ya I guess maybe they want you to plug it in completely unsimplified like this? It's important to put brackets around your f(-8) to show that it's being applied to every term over there.
i did that .-.
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well there's an incorrect 2 placed somewhere in there but i changed it and it is still wrong
You still don't have the brackets around the orange like I mentioned.
omg
you're right! i thought i had put them in lol
thank you! it comes out as correct now!
wow that's a lot of brackets <.< lol there must be a better way to do that

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