anonymous
  • anonymous
PLEASE HELP :( !! y=13e^(3x^2)-6x at x=3 in the form y=mx +b has
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
f'(x)= 78xe^(3x^2)-6
anonymous
  • anonymous
slope= 234e^27-6 y=13e^27-18
zepdrix
  • zepdrix
So you're looking for an equation of the line tangent to the curve f(x) at x=3?

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anonymous
  • anonymous
omg this is so simple
anonymous
  • anonymous
wat is america coming to
anonymous
  • anonymous
yes, Determine the equation of the line tangent to the graph of
anonymous
  • anonymous
yeah but I'm not sure why I'm getting this wrong.
zepdrix
  • zepdrix
@akotto4897 wtf is wrong with you? Why would you call someone elses problem `so simple`. It's like someone in 5th grade calling 3rd grade math SUPER EASY! Just because you understand it doesn't mean that everyone else is on your level...
zepdrix
  • zepdrix
Ok that's enough of my rant :) lemme check your work a sec :3
anonymous
  • anonymous
lol, i appreciate that. but yes, let's begin.
zepdrix
  • zepdrix
Mmm ok so your slope looks good. So what coordinate pair are we using to find our y-intercept?\[\Large f(3)\quad=\quad 13e^{27}-18\]Mmm ok ok I see you have the written down already :3 \[\Large y_{\tan}\quad=\quad mx+b\]So if we plug in our coordinate pair, we get something like this, yes?\[\Large 13e^{27}-18\quad=\quad (234e^{27}-6)\cdot 3+b\]
anonymous
  • anonymous
woah so the slope is good and the y too.
anonymous
  • anonymous
oh i see there's an x there.
anonymous
  • anonymous
so would it be 247e^27-18?
zepdrix
  • zepdrix
for your y_tan? or for b?
anonymous
  • anonymous
y tan
zepdrix
  • zepdrix
Hmm after you find your b value, you should plug it back into the \[\Large\bf y_{tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\] For our final answer, we DON'T want the coordinate pair plugged in. We're only trying to fill in these blue pieces.
anonymous
  • anonymous
so how do i get to the answer.
anonymous
  • anonymous
at first i thought it was 247e^27 -6x
zepdrix
  • zepdrix
grr sorry website crashed on me >:c
anonymous
  • anonymous
its fine
anonymous
  • anonymous
i just don't understand what to do at the point slope formula.
anonymous
  • anonymous
i keep getting the answer wrong.
zepdrix
  • zepdrix
\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\]Plugging in our slope,\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{(234e^{27}-6)}x+\color{royalblue}{b}\]Then we plug in our coordinate pair to find b,\[\Large\bf \color{#DD4747 }{13e^{27}-18}\quad=\quad \color{royalblue}{(234e^{27}-6)}\cdot\color{#DD4747 }{3}+\color{royalblue}{b}\]
zepdrix
  • zepdrix
So we have some -18's cancelling out. And I think we get a b value of,\[\Large\bf \color{royalblue}{b\quad=\quad -689e^{27}}\]Something like that?
anonymous
  • anonymous
isn't it y-(13e^27-18)=234e^27-6x+18
zepdrix
  • zepdrix
y-?
anonymous
  • anonymous
right, y-y1=m(x-x1)
anonymous
  • anonymous
so in this case, slope= 234e^27-6 y=13e^27-18
anonymous
  • anonymous
and x1= 3
zepdrix
  • zepdrix
No that's `point-slope` form of a line. y-y1 = m( x-x1) We don't want to use that. We were told to use the `slope-intercept` form of a line. y=mx+b
anonymous
  • anonymous
can we do it the way i had it please?
anonymous
  • anonymous
yeah but i thought i could use that to find the equation of a tangent line
anonymous
  • anonymous
I've been using that for other problems and it worked.
zepdrix
  • zepdrix
Yes we can use that, so let's see if we can get it set up correctly. Lemme see if I can match what you wrote a sec :)
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
\[\Large\bf y-y_1\quad=\quad \color{royalblue}{m}(x-x_1)\]\[\Large\bf y-y_1\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-x_1)\]Then plugging in our point:\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]
anonymous
  • anonymous
yes correct! :)
anonymous
  • anonymous
after that I'm lost lol
zepdrix
  • zepdrix
Ok so now we just need to get in into the form y=mx+b. So we need to multiply out the brackets, then we gotta isolate the y term.
zepdrix
  • zepdrix
\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18\]I think it expands like that, yes? Give it a try :)
anonymous
  • anonymous
yup, but i seem to be doing that wrong unfortunately.
anonymous
  • anonymous
wait don't we just multiply out the -6 and -6*-3 only?
anonymous
  • anonymous
and leave the
zepdrix
  • zepdrix
Yes, true. We probably want don't want the x's expanded like that,\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad (234e^{27}-6)x -702e^{27}+18\]
zepdrix
  • zepdrix
\[\Large 234e^{27}\cdot-3\]
anonymous
  • anonymous
why don't we multiply 234e^27 with x?
anonymous
  • anonymous
and then -3
zepdrix
  • zepdrix
\[\large\bf (234e^{27}-6)(x-3)\quad=\\ \large\bf\quad (234e^{27}-6)x-(234e^{27}-6)3\]And we only want to multiply out the part with the 3.
anonymous
  • anonymous
im so confused. i do't understand why we did this problem differently than all the others when it comes to this part.
anonymous
  • anonymous
so for (234e^27−6)(x−3) i thought we do foil
zepdrix
  • zepdrix
Yes, foiling should give you this,\[\large\bf 234e^{27}x-6x -702e^{27}+18\]
zepdrix
  • zepdrix
Are you not getting that when you foil..?
anonymous
  • anonymous
yes i have that! :)
zepdrix
  • zepdrix
We don't want multiple x terms, so we'll factor an x out of each of the first two terms.\[\Large\bf =\quad \large\bf (234e^{27}-6)x -702e^{27}+18\]
anonymous
  • anonymous
now what? :(
anonymous
  • anonymous
where does my foiling go?
anonymous
  • anonymous
great so we factor?
zepdrix
  • zepdrix
We had this,\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]It foiled to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad 234e^{27}x-6x -702e^{27}+18\]We factored out an x to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]
zepdrix
  • zepdrix
We finish by solving for y.
anonymous
  • anonymous
right
anonymous
  • anonymous
x(-468+12)
zepdrix
  • zepdrix
\[\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]
zepdrix
  • zepdrix
what..? :(
zepdrix
  • zepdrix
the -6 and the 18 are not like terms, we can't combine those.
zepdrix
  • zepdrix
I don't know why you insisted on using point-slope form for this one. It seems way more complicated :c
anonymous
  • anonymous
but it's the only way for me… i have to stick to this or else i'll be completely lost in the exam
zepdrix
  • zepdrix
This one only ended up being complicated because our slope was `2 terms`. So it ended up being a ton of multiplication.
anonymous
  • anonymous
damn im so confused.
anonymous
  • anonymous
=[
anonymous
  • anonymous
what would be he answer btw?
zepdrix
  • zepdrix
\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]From here: you add (13e^{27}-18) to each side,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18\color{orangered}{+(13e^{27}-18)}\]Does that step make sense? +_+ That's how we isolate the y term.
zepdrix
  • zepdrix
From there we can drop the brackets since nothing in being applied to them, \[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]It looks like the 18's will cancel out,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+\cancel{18}+13e^{27}-\cancel{18}\]
zepdrix
  • zepdrix
And the exponential terms ( not the one attached to the x ) are like terms, so we'll combine them,\[\large\bf y=\\ \large\bf(234e^{27}-6)x \color{orangered}{-702e^{27}+13e^{27}}\] \[\large\bf y=(234e^{27}-6)x \color{orangered}{-689e^{27}}\]And I think that's our final answer, assuming I didn't make any boo boos along the way.
anonymous
  • anonymous
wait which is the mx and b?
anonymous
  • anonymous
oh it is correct….
zepdrix
  • zepdrix
\[\large\bf y=\color{royalblue}{(234e^{27}-6)}x \color{orangered}{-689e^{27}}\]\[\large\bf y=\color{royalblue}{m}x +\color{orangered}{b}\]

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