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mathcalculus
PLEASE HELP :( !! y=13e^(3x^2)-6x at x=3 in the form y=mx +b has
f'(x)= 78xe^(3x^2)-6
slope= 234e^27-6 y=13e^27-18
So you're looking for an equation of the line tangent to the curve f(x) at x=3?
omg this is so simple
wat is america coming to
yes, Determine the equation of the line tangent to the graph of
yeah but I'm not sure why I'm getting this wrong.
@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...
Ok that's enough of my rant :) lemme check your work a sec :3
lol, i appreciate that. but yes, let's begin.
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?\[\Large 13e^{27}-18\quad=\quad (234e^{27}-6)\cdot 3+b\]
woah so the slope is good and the y too.
oh i see there's an x there.
so would it be 247e^27-18?
for your y_tan? or for b?
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.
so how do i get to the answer.
at first i thought it was 247e^27 -6x
grr sorry website crashed on me >:c
i just don't understand what to do at the point slope formula.
i keep getting the answer wrong.
\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\]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,\[\Large\bf \color{#DD4747 }{13e^{27}-18}\quad=\quad \color{royalblue}{(234e^{27}-6)}\cdot\color{#DD4747 }{3}+\color{royalblue}{b}\]
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?
isn't it y-(13e^27-18)=234e^27-6x+18
right, y-y1=m(x-x1)
so in this case, slope= 234e^27-6 y=13e^27-18
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
can we do it the way i had it please?
yeah but i thought i could use that to find the equation of a tangent line
I've been using that for other problems and it worked.
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 :)
\[\Large\bf y-y_1\quad=\quad \color{royalblue}{m}(x-x_1)\]\[\Large\bf y-y_1\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-x_1)\]Then plugging in our point:\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]
after that I'm lost lol
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.
\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18\]I think it expands like that, yes? Give it a try :)
yup, but i seem to be doing that wrong unfortunately.
wait don't we just multiply out the -6 and -6*-3 only?
Yes, true. We probably want don't want the x's expanded like that,\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad (234e^{27}-6)x -702e^{27}+18\]
\[\Large 234e^{27}\cdot-3\]
why don't we multiply 234e^27 with x?
\[\large\bf (234e^{27}-6)(x-3)\quad=\\ \large\bf\quad (234e^{27}-6)x-(234e^{27}-6)3\]And we only want to multiply out the part with the 3.
im so confused. i do't understand why we did this problem differently than all the others when it comes to this part.
so for (234e^27−6)(x−3) i thought we do foil
Yes, foiling should give you this,\[\large\bf 234e^{27}x-6x -702e^{27}+18\]
Are you not getting that when you foil..?
yes i have that! :)
We don't want multiple x terms, so we'll factor an x out of each of the first two terms.\[\Large\bf =\quad \large\bf (234e^{27}-6)x -702e^{27}+18\]
where does my foiling go?
great so we factor?
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,\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]
We finish by solving for y.
\[\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]
the -6 and the 18 are not like terms, we can't combine those.
I don't know why you insisted on using point-slope form for this one. It seems way more complicated :c
but it's the only way for me… i have to stick to this or else i'll be completely lost in the exam
This one only ended up being complicated because our slope was `2 terms`. So it ended up being a ton of multiplication.
damn im so confused.
what would be he answer btw?
\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]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.
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,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+\cancel{18}+13e^{27}-\cancel{18}\]
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.
wait which is the mx and b?
oh it is correct….
\[\large\bf y=\color{royalblue}{(234e^{27}-6)}x \color{orangered}{-689e^{27}}\]\[\large\bf y=\color{royalblue}{m}x +\color{orangered}{b}\]