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Math homework help

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Flacker
Posts: 584 - Karma: 1,329
So I need some help with my math homework. I need to find the slope of the line tangent to the given point y=2x^2 at (-2,8) I know your supposed to find the derivative but I don't know what to so from there
Nov 20 2013 8:22PM
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Wallisch_pl
Posts: 5260 - Karma: 1,022
I just had to go find my algebraic calc, lol, and I don't understand your question,
Nov 20 2013 8:25PM
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chickenonabun
Posts: 461 - Karma: 285
So using the power rule you find the derivative (which is its own function that describes the slope of the first function at any giver point. The derivative is 4x. Then you put The x value you want to find the slope of in for the derivative, and you get f'(-2)=-8.
Nov 20 2013 8:28PM
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proZach
Posts: 7163 - Karma: 919
http://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version
Nov 20 2013 8:28PM
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GIF
Posts: 468 - Karma: 1,279
No one cares!!
Nov 20 2013 8:29PM
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Miomo
Posts: 4302 - Karma: 14,133
You could always just skip the problem. Seeing as it's homework.
Nov 20 2013 8:46PM
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Flacker
Posts: 584 - Karma: 1,329
I figured it out actually but now I can't remember how to find the critical point of a function
Nov 20 2013 8:54PM
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J-T
Posts: 269 - Karma: 697
Yeah we just did this in Calc like a month ago, this guy's got it spot on.
Nov 20 2013 8:55PM
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jahmin44
Posts: 638 - Karma: 37
Critical point of a function is a maximum or minimum point or its graph, at this point the slope has to be zero. So you find the x value were the derivative = 0. f(x)= 2x^2, f'(x)=4x. so the critical point is at x = 0. Which makes sense because your function is just a parabola.
Nov 20 2013 9:18PM
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SlurmrnHurnyBurn
Posts: 315 - Karma: 765
if you can't figure out the slope of a line you're gonna have some troubles with the rest of life.
Nov 20 2013 9:19PM
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Theskiingguy
Posts: 476 - Karma: 2,514
it's a tangent line of an exponential equation/function... not a linear line, totally different bud.

But yeah, that guy a couple posts down is right. But if they're making you do the
f(x-a) - f(a)
--------------
f(a)

or whatever that equation is, then that's totally different. Let me know if thats what you're supposed to do and I can help. Just can't remember the exact equation
Nov 20 2013 9:39PM
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coling
Posts: 1086 - Karma: -3
Yup
Nov 20 2013 9:47PM
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Captain_Slow
Posts: 1677 - Karma: 24
I dont think he'll be using that method to find the answer... i learned that in precal but skipped it in calc (which im assuming hes taking now)
Nov 20 2013 11:23PM
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Theskiingguy
Posts: 476 - Karma: 2,514
We had to do it in our calc class in the beginning, but then we were able to use the "short cut" way, so idk
Nov 20 2013 11:51PM
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adzy
Posts: 151 - Karma: 370
Lol I wish I could go back in time when calc was this easy.
Nov 21 2013 12:04AM
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DaffyDaDak
Posts: 864 - Karma: 433
Bump so I dont have to make a new thread,
this projects due tomorrow and ive done math all day and im exhausted and cant get my mind to compute anymore can someone help me with this.

Write an equation (y = mx+b) for the graph of a linear function that:
has slope 5 and y-intercept -5
has slope -5/6 and y-intercept -6/5
has slope -5 and y-intercept 6
Dec 9 2015 1:40PM
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erikK
Posts: 3245 - Karma: 2,328
DaffyDaDakBump so I dont have to make a new thread,
this projects due tomorrow and ive done math all day and im exhausted and cant get my mind to compute anymore can someone help me with this.

Write an equation (y = mx+b) for the graph of a linear function that:
has slope 5 and y-intercept -5
has slope -5/6 and y-intercept -6/5
has slope -5 and y-intercept 6


m is the slope and b is the y intercept

oh to be in grade 9 again...
Dec 9 2015 1:45PM
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itzki
Posts: 444 - Karma: 1,344
DaffyDaDakBump so I dont have to make a new thread,
this projects due tomorrow and ive done math all day and im exhausted and cant get my mind to compute anymore can someone help me with this.

Write an equation (y = mx+b) for the graph of a linear function that:
has slope 5 and y-intercept -5
has slope -5/6 and y-intercept -6/5
has slope -5 and y-intercept 6


erikKm is the slope and b is the y intercept

oh to be in grade 9 again...


A.k.a...

y=5x-5

y=-5/6x-6/5

y=-5x+6
Dec 9 2015 1:49PM
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DaffyDaDak
Posts: 864 - Karma: 433
erikKm is the slope and b is the y intercept

oh to be in grade 9 again...


itzkiA.k.a...

y=5x-5

y=-5/6x-6/5

y=-5x+6


grade 11 actually haha.... i got another one for you guys
For the function f(x) = -3x + 8, determine f(-2).
Dec 9 2015 4:02PM
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DaffyDaDak
Posts: 864 - Karma: 433
DaffyDaDakgrade 11 actually haha.... i got another one for you guys
For the function f(x) = -3x + 8, determine f(-2).


pretty sure its 2
Dec 9 2015 4:07PM
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AgitatedHiatus
Posts: 319 - Karma: 1,053
DaffyDaDakpretty sure its 2


It's not. Since f(x) means you put whatever number you have as "x" in the equation, -2 is taking the place of x in the equation. You take -3(-2) + 8 which gives you an answer of 14.
Dec 9 2015 4:35PM
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lucarichardson
Posts: 85 - Karma: 35
Step 1. Use power rule to find derivative of the function 2x^2:
Constant 2 times n (in x^n) times x^(n-1), or x^1, or x
fprime= 4x
Step 2. Plug in x value -2 to find the derivative at that point.
fprime(-2)=4(-2)
fprime(-2)= -8
Step 3. Pee in her butt
Step 4. Slope of the tangent line at (-2, 8)= fprime(-2)= -8
Step 5. Find tangent line using point slope form
y= -8(x+2)+8
Dec 12 2015 7:39PM
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Deepskier
Posts: 3267 - Karma: 4,829
erikKm is the slope and b is the y intercept

oh to be in grade 9 again...


you mean grade 6
Dec 13 2015 9:21PM
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