PLEASE HELP :( !!
y=13e^(3x^2)-6x
at x=3 in the form y=mx +b has

- anonymous

PLEASE HELP :( !!
y=13e^(3x^2)-6x
at x=3 in the form y=mx +b has

- jamiebookeater

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- anonymous

f'(x)= 78xe^(3x^2)-6

- anonymous

slope= 234e^27-6
y=13e^27-18

- zepdrix

So you're looking for an equation of the line tangent to the curve f(x) at x=3?

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## More answers

- anonymous

omg this is so simple

- anonymous

wat is america coming to

- anonymous

yes, Determine the equation of the line tangent to the graph of

- anonymous

yeah but I'm not sure why I'm getting this wrong.

- 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

Ok that's enough of my rant :) lemme check your work a sec :3

- anonymous

lol, i appreciate that. but yes, let's begin.

- 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

woah so the slope is good and the y too.

- anonymous

oh i see there's an x there.

- anonymous

so would it be 247e^27-18?

- zepdrix

for your y_tan? or for b?

- anonymous

y tan

- 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

so how do i get to the answer.

- anonymous

at first i thought it was 247e^27 -6x

- zepdrix

grr sorry website crashed on me >:c

- anonymous

its fine

- anonymous

i just don't understand what to do at the point slope formula.

- anonymous

i keep getting the answer wrong.

- 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

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

isn't it y-(13e^27-18)=234e^27-6x+18

- zepdrix

y-?

- anonymous

right, y-y1=m(x-x1)

- anonymous

so in this case, slope= 234e^27-6
y=13e^27-18

- anonymous

and x1= 3

- 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

can we do it the way i had it please?

- anonymous

yeah but i thought i could use that to find the equation of a tangent line

- anonymous

I've been using that for other problems and it worked.

- 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

ok

- 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

yes correct! :)

- anonymous

after that I'm lost lol

- 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

\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18\]I think it expands like that, yes?
Give it a try :)

- anonymous

yup, but i seem to be doing that wrong unfortunately.

- anonymous

wait don't we just multiply out the -6 and -6*-3 only?

- anonymous

and leave the

- 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

\[\Large 234e^{27}\cdot-3\]

- anonymous

why don't we multiply 234e^27 with x?

- anonymous

and then -3

- 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

im so confused. i do't understand why we did this problem differently than all the others when it comes to this part.

- anonymous

so for (234e^27−6)(x−3) i thought we do foil

- zepdrix

Yes, foiling should give you this,\[\large\bf 234e^{27}x-6x -702e^{27}+18\]

- zepdrix

Are you not getting that when you foil..?

- anonymous

yes i have that! :)

- 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

now what? :(

- anonymous

where does my foiling go?

- anonymous

great so we factor?

- 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

We finish by solving for y.

- anonymous

right

- anonymous

x(-468+12)

- zepdrix

\[\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]

- zepdrix

what..? :(

- zepdrix

the -6 and the 18 are not like terms, we can't combine those.

- zepdrix

I don't know why you insisted on using point-slope form for this one.
It seems way more complicated :c

- 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

This one only ended up being complicated because our slope was `2 terms`.
So it ended up being a ton of multiplication.

- anonymous

damn im so confused.

- anonymous

=[

- anonymous

what would be he answer btw?

- 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

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

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

wait which is the mx and b?

- anonymous

oh it is correct….

- 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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